答:
In this case $x \in S^N$ where $S$ is some set means that $x$ is a tuple of length $N$ with entries from the set $S$, that is

$$x = (s_1, s_2, \ldots, s_N)$$

where all the $s_i$ are elements of $S$. So in the case of $S = \{0, 1\}$ (the set containing the elements $0$ and $1$) and $N=2$ we have $$S^N= \{(0,0),(0,1),(1,0),(1,1)\}.$$

In the case of $S = \mathbb R_{>0}$, the entries are positive numbers.

Title: How to interpret / read the following mathematical notations?

Body:
In one of the paper I am reading, I came across two mathematical notations which I cannot understand. The author also did not give any explanation. In both, note that $N$ is a positive integer.

Term 1: $x \in \{ 0,1 \}^N$

Term 2: $h\in \mathbb{R}^N_{>0}$

Thanks


问:
Cancelling terms in polynomial congruence
If we have a polynomial like $f(x)=x^{18}+7x^{13}-2x^5+8$. If we want to consider it mod 5 for example, can we simplify it like by subtracting terms (or adding multiples) in 5's; for example, is $f(x)$ equal to $x^{18}+2x^{13}+3x^5+3$ mod 5? I think I've seen this done before but it feels like this shouldn't be possible. And if it is (or isn't), why is that?

答:
Yes, this is perfectly fine. In you example, because $5\equiv0\pmod{5}$, also $5x^5,5x^{13}\equiv0\pmod{5}$, and hence
\begin{eqnarray*}
f(x)&=&x^{18}+7x^{13}-2x^5+8\\&\equiv&(x^{18}+7x^{13}-2x^5+8)+(-5x^{13}+5x^5-5)\\
&=&x^{18}+2x^{13}+3x^5+3\pmod{5}.
\end{eqnarray*}
Similarly, as suggested in the comments, the difference between the two polynomials equals
$$-5x^{13}+5x^5-5=5(-x^{13}+x^5-1),$$
which is a multiple of $5$. So the two polynomials are congruent modulo $5$.

ContentLicense: CC BY-SA 4.0


问:
Simplest (completely) positive maps
So, this question may be very simple, but I have an element $a$ in a C*-algebra, we can assume $a\in M_n(\mathbb{C})$ and therefore
$$ a = b a b^{-1} $$
has the same spectrum. Therefore, $x\mapsto bxb^{-1}$ maps positive elements to positive elements. But on the other hand the simplest completely positive map is $x\mapsto cxc^*$ and I don't get now why this is a positive map or where is the mistake.

答:
If $a$ is positive and $b$ is invertible, then $bab^{-1}$ is not necessarily positive. The problem is that for an element to be positive it does not suffice that its spectrum is contained in $[0,\infty)$, it additionally has to be self-adjoint. For example, the matrix
$$
\begin{pmatrix}0&1\\0&0\end{pmatrix}
$$
has spectrum $\{0\}$, but it is not positive (it's clearly not self-adjoint).

Title: Simplest (completely) positive maps


问:
If $\sup_n P(\vert X_n -a_n\vert > b) \leq c$, then can I find a $d \in \mathbb{R}$ such that $\sup_n P(\vert X_n \vert > d) \leq c$?
Let $\{X_n\}_{n \in \mathbb{N}}$ denote a sequence of real random variables and $\{a_n\}_{n \in \mathbb{N}}$ a sequence of real numbers. Suppose $P(\vert X_n\vert>\epsilon) \to 0$ for $n \to \infty$ for all $\epsilon>0.$

If $\sup_n P(\vert X_n -a_n\vert > b) \leq c$, then can I find a $d \in \mathbb{R}$ such that $\sup_n
 P(\vert X_n \vert > d) \leq c$? And how to choose $d$?

So I know $X_n$ converges to $0$ in probabillity. Can I use that it is a deterministic limit to say something about convergence almost surely? Or can I use
$$
\sup_n P(\vert X_n \vert  > b+ \vert a_n\vert) = \sup_n P(\vert X_n \vert - \vert a_n\vert > b) \leq \sup_n P(\vert X_n -a_n\vert > b) \leq c
$$
somehow?

答:
Okay I think Davide answers my question but I do have a follow up question. What if instead the setup was:

Let $\{X_{nk} \mid n \in \mathbb{N}, k \leq n \}$ denote a traingular array of real random variables and $\{a_n\}_{n \in \mathbb{N}}$ a sequence of real numbers. Suppose $\max_k P(\vert X_{kn}\vert>\epsilon) \to 0$ for $n \to \infty$ for all $\epsilon>0.$

If $\sup_n \sum_{k=1}^n P(\vert X_{kn} -a_n\vert > b) \leq c$, then can I find a $d \in \mathbb{R}$ such that $\sup_n
 \sum_{k=1}^n  P(\vert X_{nk} \vert > d) \leq c$? And how to choose $d$?

As far as I can tell you need a different solution since this with $d=\max\{1,d_1,\dots,d_{n_0 -1}\}$ would yield the bound
$$
\sum_{k=1}^n  P(\vert X_{nk} \vert > d) \leq nc
$$
which doesn't give anything useful when applying $\sup_n$ on both sides.


问:
Find $g$ such that $\operatorname{div}g=f$ in the whole space $\mathbb{R}^d$. (Bogovskii operator lemma)
For a fixed $f \in L^2(Q)$ with $Q$ a bounded domain of $\mathbb{R}^d$, I know we can find a function $g \in H^1_0(Q)$ such that

$$\operatorname{div}g =f \enspace\text{ in } Q$$
in the sense of distribution. This is only possible if we assume the compatibility condition :

$$\int_Q f =0 \quad \quad \quad (\star)$$

This theorem is also known as the Bogovskii operator, and there exist milder versions (with $f \in L^p(Q)$ essentially) that are well referenced in the book of Galdi, in the dedicated chapter :

Galdi, G. (2011). An introduction to the mathematical theory of the Navier-Stokes equations: Steady-state problems.

I was wondering if the following is true :

Let $f \in L^2(\mathbb{R}^d)$, can we find $g$ (for exemple in $H^1(\mathbb{R}^d)$) such that $\operatorname{div}g =f$ in $\mathbb{R}$ ? Is there a compatibility condition like $(\star)$ ?

Any references or help is welcomed.

答: If you assume that $f \in L^{\frac{2d}{d+2}}(\mathbb{R}^d) \cap L^2((\mathbb{R}^d)$ then the answer is yes (for some $g \in L^2((\mathbb{R}^d)$  : see Lemma 1.6.2 Chapter II in the book of Sohr - The Navier-Stokes equations: An elementary functional analytic approach.


问:
Find the rejection region of a random sample of exponential distribution
I'd like help with the following problem:

Consider a random sample $X_1, ..., X_n$ from the Exponential distribution $\epsilon(\theta)$ with density $\theta e^{-\theta x}$, $x>0$. Suppose we wish to test the hypothesis $H_0: \theta=1$ vs $H_1: \theta \neq 1$.

(1) Calculate the likelihood ratio statistic for this testing problem and (2) show that the LRT is based on the statistic $T=\sum_{i=1}^nX_i$ and rejects if $T\in(0,c_1]\cup[c_2,\infty)$ for some suitably chosen $c_1, c_2$.

My work in a nutshell: Knowing that the likelihood function is given by $L(\theta;x_1,...,x_n)=\theta^ne^{-\theta n\bar{x}}$ and that the likelihood ratio statistic is given by $\lambda(x_1,...,x_n)=\frac{\sup_{\Theta=1}{L(\theta;x_1,...,x_n)}}{\sup_{\theta\in\Theta}L(\theta;x_1,...,x_n)}$, I find the MLE estimator to be $\hat{\theta}=\frac{1}{\bar{x}}$, plug it into the denominator, ending up with $\lambda(x_1,...,x_n)=\bar{x}^n e^{-n\bar{x}+1}=(\frac{\sum_{i=0}^n X_i}{n})^ne^{-\sum_{i=0}^n X_i+n}$, which I believe is enough to show that the LRT is based on $T=\sum_{i=0}^nX_i$. Now, knowing that the rejection region is given by $R=\{T:\lambda\leq c\}$ for some set critical value of $c$ (chosen as a function of $\alpha$). I then end up with $(\frac{\sum_{i=0}^n X_i}{n})^ne^{-\sum_{i=0}^n X_i+n}\leq c$, which I'm not sure is correct (and would like confirmation before I proceed with straight computation to find the two rejection regions $(0,c_1]\cup[c_2,\infty)$.

Thanks a bunch for your time. For context, I'm a first-year math major.

答:
I then end up with ...

more or less right, the sum goes from 1 to n, not from zero. The $n$ in the exponent is obviously out of the summation... in a more suitable notation, you get that

$$\lambda(\mathbf{x})\propto T^n e^{n-T}$$

Where $T=\Sigma_iX_i\sim \text{Gamma}(n;\theta)$

Doing a drawing of your likelihood ratio (I fixed some $n$ to have the graphics) you see this

thus the probability for $\lambda(\mathbf{x})<c$ is equivalent to $T\leq c_1$ or $T\geq c_2$ for appropriately chosen constants $c_1,c_2$

Of course the two quantiles $c_1,c_2$ can be derived by a $\chi_{(2n)}^2$ distribution and consequently found using the standard table


问:
Independence of $\sigma-$fields
Let $(\Omega, \mathcal{F}, P)$ be a probability space. I have two questions:

Is is true that if random variable $X$ is independent from $\sigma-$fields $\mathcal{A},\mathcal{B}\subset \mathcal{F}$, then $X$ is independent from $\sigma-$field $\sigma \left(\mathcal{A}\cup \mathcal{B}\right)$?

Let $N=\{Y\subset Z:P(Z)=0\}$. Is it true that $\sigma(\mathcal{A}\cup N)=\sigma(\mathcal{A}\cup\sigma(N))$?

答:
To your questions:

$X$ is said to be independet from $\mathcal{A},\mathcal{B}\subset \mathcal{F}$ if the $\sigma$-algebra $\sigma(X)$ generated by $X$ is independent from the $\sigma$-algebras $\mathcal{A},\mathcal{B}$. This means for all $A\in\mathcal{A}$, $C\in\sigma(X)$ the equation $P(A\cup C)=P(A)P(C)$ holds, likewise for $B\in\mathcal{B}$. If you can guarantee that $\mathcal{A}\cup \mathcal{B}$ is closed under finitely many intersections, then you only have to check that for all $M\in\mathcal{A}\cup \mathcal{B}$ and $C\in\sigma(X)$ the equation $P(M\cup C)=P(M)P(C)$ holds, which is trivial, as $M\in\mathcal{A}\cup \mathcal{B}$ implies $M\in\mathcal{A}\lor M\in\mathcal{B}$. Without the additional assumption, I think it is not true in general.

Frist we have to clearify the set $N:=\{Y\in\mathcal B|\exists Z\in\mathcal B: Y\subseteq Z\land P(Z)=0\}$. As $\mathcal{A}\cup N\subseteq \mathcal{A}\cup\sigma(N)$ and the $\sigma$-operator is isotone, we get $\sigma(\mathcal{A}\cup N)\subseteq\sigma(\mathcal{A}\cup\sigma(N))$. For the opposite direction: From isotonicity follows further $\sigma(N)\subseteq \sigma(\mathcal{A}\cup N)$, $\mathcal{A}\subseteq \sigma(\mathcal{A}\cup N)$ and thus $\mathcal{A}\cup\sigma(N)\subseteq \sigma(\mathcal{A}\cup N)$. As the $\sigma$-operator creates the smalest $\sigma$-algebra containing the set applied on, $\sigma(\mathcal{A}\cup\sigma(N))\subseteq \sigma(\mathcal{A}\cup N)$ follows.

Title: Independence of $\sigma-$fields


问:
Evaluating $\int_{0}^{1} \frac{ \ln x \ln (1-x)}{\sqrt{x} \sqrt{1-x}} dx$
I have the following integral:
$$\int_{0}^{1} \frac{ \ln x \ln (1-x)}{\sqrt{x} \sqrt{1-x}} dx$$

I think I may be to evaluate this with beta and gamma functions but I am not quite sure how. Any help?

答:
Let $I$ denote the integral in question. With the substitution $x = \sin^{2} \theta$, we have

$$ I = 8 \int_{0}^{\frac{\pi}{2}} \log \cos \theta \log \sin \theta \, d \theta = \frac{\partial^2}{\partial z \partial w} \beta \left( \frac{1}{2}, \frac{1}{2} \right), $$

where

$$\beta(z, w) = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)} = 2 \int_{0}^{\frac{\pi}{2}} \cos^{2z-1}\theta \sin^{2w-1}\theta \, d\theta \tag{1} $$

is the beta function. This gives

$$ I = 4\pi \log^{2} 2 - \frac{\pi^3}{6}. $$

ContentLicense: CC BY-SA 3.0


问:
Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$
I tried to re-arrange the terms

$6x^2 + 2xy + 6y^2 = 9 $

$6x^2 + 6y^2 = 9 - 2xy $

$6 (x^2 + y^2) = 9 - 2xy $

$x^2 + y^2 = \frac{9 - 2xy}{6} $

Using A.M $\geq$ G.M

$\frac{x^2 + y^2}{2} \geq xy $

Can ayone help me from here? Am I going correct?

答:
Let set: $\begin{cases}f(x,y)=x^2+y^2\\g(x,y)=6x^2+2xy+6y^2\end{cases}$

The curve $(\mathcal E): g(x,y)=9$ is an ellipse centered at the origin with major axis supported by $y=-x$ and minor axis supported by $y=x$.

Indeed if you calculate its equation in the $45^\circ$ rotated basis then you get it in its reduced form:

$$g(\tfrac{X+Y}{\sqrt{2}},\tfrac{X-Y}{\sqrt{2}})=7X^2+5Y^2=9$$

On the other hand, the curve $(\mathcal C): f(x,y)=k$ is just a circle centered at origin.

The linked extrema occurs when this circle is tangent to the ellipse, therefore the minimum occurs on the minor axis (i.e. when $x=y$) and the maximum occurs on the major axis (i.e. when $x=-y$).

$\begin{cases}g(x,x)=14x^2=9\\g(x,-x)=10x^2=9\end{cases}\implies \begin{cases}f(x,x)=2x^2=\frac 97&\text{min}\\f(x,-x)=2x^2=\frac 95&\text{max}\end{cases}$

Note: you can also extrapolate these values directly from the reduced equation of the ellipse.


问:
What does "$\imath$" (a cursive "i" without a tittle) mean?
What does "$\imath$" (a cursive "i" without a tittle) mean?

It seems to be an algebraic symbol, as that was all I was dealing with in the original question.

Here's the original question (I was trying to simplify it and was using cymath.com to check):

$$v = \pm\sqrt{-c^2\left(\frac{t^2}{x^2}-1\right)}\qquad\text{Simpler Form:}\quad v = \pm\sqrt{c^2\left(\frac{t^2}{x^2}-1\right)\imath}$$

Does anyone know what this symbol means? Thanks in advance.

答:
The symbol \iota $\iota$ (Greek letter iota) is used in some countries as $\sqrt{-1}$.  Others use $i$ or ${\rm i}$ or $j$. The other symbol \imath $\imath$ is not intended for this, instead it is used when $i$ has some decoration above it: $\vec{\imath}$ looks better than $\vec{i}$ .

The two are slightly different.  Enlarged view:

Also, $i$ is not "cursive" it is "italic".  Here is "cursive", also known as "script".  But you can do this only if you have the proper fonts installed.


问:
How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$?
How can the following inequation be proven?

$$a^2 + b^2 + c^2 \ge ab + bc + ca$$

答: Schur inequality where $r= 0.$

Body: Schur inequality where $r= 0.$


问:
Proving that a wf is logically valid/not valid.
This 2 question are from "Introduction to Mathematical Logic" by Elliot Mendelson , forth edition , exercise 2.21(g) and 2.22(i).

2.21.g. ($\exists x)(\forall y)(A^2_1(x,y) \land \lnot A^2_1(y,x) \to [A^2_1(x,x) \leftrightarrow A^2_1(y,y)])$

I have to prove that this formula is not logically valid with a counterexample.Here are the counterexamples I was able to come up with so far:

$(A^2_1)^M(x,y) \equiv x = y$

$(A^2_1)^M(x,y) \equiv x \neq y$

$(A^2_1)^M(x,y) \equiv x < y$

$(A^2_1)^M(x,y) \equiv x > y$

$(A^2_1)^M(x,y) \equiv x \geq y$

$(A^2_1)^M(x,y) \equiv x \leq y$

But none of them seem to work though and I am all out of ideas.

This is 2.22.i.

For all positive integers $x$, $x \leq x$.For all positive integers $x$,$y$,$z$, if $x \leq y$ and $y \leq z$, then $x \leq z$. For all positive integers $x$ and $y$ , $x \leq y$ or $y \leq x$.Therefore, there is a positive integer $y$ such that, for all positive integers $x$, $y \leq x$.

Lets say $x\leq y \equiv A(x,y)$

There are three premises (1,2,3) and one conclusion (4).

$\forall x [A(x,x)]$

$\forall x \forall y \forall z [A(x,y) \land A(y,z) \to A(x,z)]$

$\forall x \forall y [A(x,y) \lor A(y,x)]$

$\exists y \forall x [A(y,x)]$

I have to check whether the conjunction of the premises logically imply the conclusion.

My first confusion is , why do we need (1.) here seperately ? For example , from (3.)

$\to \forall x \forall y [A(x,y) \lor A(y,x)]$

$\to \forall x \forall x [A(x,x) \lor A(x,x)]$

$\to \forall x [A(x,x) \lor A(x,x)]$

$\to \forall x [A(x,x)]$

Anyway , for the main problem , I thought that it might be logically implying . So I assumed the conclusion to be false , then prove one of the premises to be wrong and show a contradiction.But I wasn't able to. Maybe it is not logically implying ?

I am all out of ideas , can someone give me a hint?

答:
For 2.21(g), define a two-elements domain $D = \{ 0,1 \}$ with a relation $A^D$ such that $(0,0), (0,1) \in A^D$.

Consider now $(∀y)(A(0,y) ∧ ¬A(y,0) \to [A(0,0) ↔ A(y,y)])$. We have two possible choices for $y$: $0$ and $1$.

For $0$ we have that the antecedent is false (due to $\lnot A(0,0)$) and thus the conditional is true.

But for $1$ we have that the antecedent is true while the consequent is false.

Thus, it is not true that the formula is true for every element in the domain $D$.

For 2.22(i), consider the set $\mathbb Z$ of integers: the three axioms are satisfied but there is no "minimum".


问:
Theorem 2.14 in Rudin's RCA: What is meant by "(a) forces (b)"?
Theorem 2.14 is the Riesz Representation Theorem(about functionals on $C_c(X)$):

Theorem 2.14. Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a linear functional on $C_c(X)$. Then there exists a $\sigma$-algebra $\mathfrak M$ in $X$ which contains all Borel sets in $X$, and there exists a unique positive mesaure $\mu$ on $\mathfrak M$ which represents $\Lambda$ in the sense that

(a) $\Lambda f = \int_X fd\mu$ for every $f \in C_c(X)$,

and which has the following additional properties:

(b) $\mu(K)<\infty$ for every compact set $K\subset X$.

(......the left omitted)

And a few lines later the author writes:

$$ \mu_1(K) = \int_X\chi_K\,d\mu_1 \le \int_X f\,d\mu_1 = \Lambda f = \int_X f\,d\mu_2 \le \int_X \chi_V\,d\mu_2 = \mu_2(V) < \mu_2(K) + \epsilon. $$
Incidentally, the above computation shows that (a) forces (b).

"(a) forces (b)" is the problem. How this is possible? The author has just used (b) to show the above inequality, and says that (a) implies (b) by the above computation. I need explanation about this.

答: If we know $\Lambda f = \int_X fd\mu$ for some $\mu$, then if $K$ is compact we can find some function $g$ that is $1$ on $K$ and compactly supported on $X$ and then this integral is at least $\mu(K)$ and must also be finite, as $\Lambda(f)$ is. So $\mu$ is then always be finite on compact sets. So for $(a)$ to hold at all, $(b)$ must be satisfied for the $\mu$ we want. I think that's all he's saying.


问:
What Is the Motivation of Proximal Mapping / Proximal Operator?
For a convex function $h$, its proximal operator is defined as:
$$\operatorname{prox}_h(x)=\arg\min_u \Big(h(u)+\frac{1}{2}\|u-x\|^2\Big)$$
Can anyone provide an intuitive explanation/motivation of proximal mapping?

答:
There are many good answers here already, but for the sake of completeness I will add on more intuition. The proximal operator of $g$ (with parameter $\lambda$) can also be seen as a gradient step (with stepsize $\lambda$) with respect to the Moreau envelope $g_\lambda$ of $g$. The Moreau envelope is a smooth under approximation of the original function and is given by
$$
g_\lambda(x) = \min_u \{g(u) + \frac{1}{2 \lambda} || u-x||^2\}.
$$
Evidently its definition is closely related to the proximal operator. The above mentioned identity then reads as
$$
\operatorname{prox}_{\lambda g}(x) = x - \lambda \nabla g_\lambda(x).
$$


问:
Why the order of elements of my group $\mathbb{Z}\left[\sqrt 7\right]$ under multiplication modulo prime $p$ often divide $p^2-1$
Consider the group $G$ with elements of the form $a + b \sqrt 7$ ($a, b \in \mathbb{Z}$) under multiplication mod $p$ where $p$ is prime.

I noticed the order of $1 + \sqrt 7$ in my group is usually of the form $\frac{p^2 - 1}{d}$ where $d$ is some positive integer divisor.

For example, operating under mod $p := 5$: $5^2 - 1 = 24$ and $\mathrm{ord}\left(1 + \sqrt 7\right) = 12 = 24/2$.

I plotted the order of $1 + \sqrt 7$ under different values of $p$ and here are the results:

You can easily see the patterns emerge (i.e. broken parabolas in the plot):

$p^2 - 1$

$(p^2 - 1)/2$

etc.

Why does this happen? Is there a reliable way to predict when this happens and even what the divisor would be? This could potentially provide an efficient way to calculate the order of my group element.

答:
Let $p$ be a prime number and $G_p$ the group you are considering. For $G_p$ to be a group in the first place, every element of the form $a+b\sqrt{7}$ must be invertible mod $p$. In particular you cannot have $a,b\equiv0\pmod{p}$.

Let's make the situation a bit more precise by working over the finite field $\Bbb{F}_p$ of $p$ elements. Adjoining $\sqrt{7}$, a root of $X^2-7\in\Bbb{F}_p[X]$, then yields a field extension $\Bbb{F}_p\subset\Bbb{F}_p[\sqrt{7}]$, and its degree is either $1$ or $2$. The degree is $1$ if and only if $X^2-7\in\Bbb{F}_p[X]$ has a root in $\Bbb{F}_p$, or equivalently, if $7$ is a quadratic residue mod $p$. By quadratic reciprocity it follows that the degree is $1$ if and only if
$$p\equiv1,2,3,7,9,25,27\pmod{28}.$$
In this case $\Bbb{F}_p[\sqrt{7}]=\Bbb{F}_p$ with unit group $\Bbb{F}_p^{\times}=\Bbb{F}_p-\{0\}$, so the invertible elements form a group of order $p-1$. Otherwise $\Bbb{F}_p[\sqrt{7}]\cong\Bbb{F}_{p^2}$, and the invertible elements form a group of order $p^2-1$. Note that in either case this group is cyclic; see this question for a proof.

As far as I can tell there is no quick method to determine the order of an element. A closely related problem is finding a primitive root mod $p$, which is hard in general. But here are a few simple observations:

If $7$ is a quadratic residue mod $p$, then the order of $a+b\sqrt{7}$ divides $p-1$.

If $b\equiv0\pmod{p}$ then $a+b\sqrt{7}\in\Bbb{F}_p$ and so the order divides $p-1$.

If $7$ is not a quadratic residue and $b\not\equiv0\pmod{p}$, then the order of $a+b\sqrt{7}$ does not divide $p-1$. Moreover $(a+b\sqrt{7})^{p+1}\in\Bbb{F}_p^{\times}$, so you can first find a divisor $d$ of $p+1$ such that $(a+b\sqrt{7})^d\in\Bbb{F}_p^{\times}$, and then use point $1$. Note that $\gcd(p-1,p+1)=2$ so you may need to take out a factor $2$.

Another approach might be to consider that $1+\sqrt{7}$ is a root of $X^2-2X-6$. Then for each prime number $p$, you want to find the minimal positive integer $d$ such that $X^2-2X-6$ divides $X^d-1$, or better yet the $d$-th cyclotomic polynomial $\Phi_d$, as polynomials in $\Bbb{F}_p[X]$. The ideas above already give some restrictions on $d$.


问:
Why is a slab convex?
A set of the form $$\left\{x \in \Bbb R^n\mid \alpha \leq a^T x \leq \beta \right\}$$ is called a slab. A slab is an intersection of two halfspaces. Hence, it is a convex set.

Can someone explain why a slab is an intersection of two halfspaces? I can't understand it.

答:
A slab is a set of the form
$$S = \{x\in\mathbb R^n : \alpha\le a^{T} x\le\beta\}$$
for some fixed $a = (a_1,a_2,\ldots,a_n)\in\mathbb R^n$ and $\alpha,\beta\in\mathbb R$.

$S$ is the intersection of $$H_1 = \{x\in\mathbb R^n : a^{T} x\ge\alpha\}$$ and $$H_2 = \{x\in\mathbb R^n : a^{T} x\le\beta\}$$
i.e. $S = H_1\cap H_2$. $H_1$ and $H_2$ are halfspaces. Hence, $S$ is convex.

Title: Why is a slab convex?


问:
How Do You Expand and Simplify?
How do you expand and simplify $(x + 2y)^2$ or $(7x + 5y)^2$. I don't understand, how would you multiply that by itself?

I mean I can simplify and expand two brackets; e.g. $(x + 3) (x + 5)$ but I'm not sure how to do one bracket with a squared.

Title: How Do You Expand and Simplify?

Body:
Expand, distribute, simplify.

$$(x+2y)^2=(x+2y)(x+2y)=x^2+2xy+2xy+4y^2=\boxed{x^2+4xy+4y^2}$$

$$(7x+5y)^2=(7x+5y)(7x+5y)=49x^2+35xy+35xy+25y^2=\boxed{49x^2+70xy+25y^2}$$


问:
Solve the differential equation ${y}'\left ( x+ y^{2} \right )= y$

Solve the differential equation
$${y}'\left ( x+ y^{2} \right )= y$$

by jp.Wolfram|Alpha _the result is $\left ( 2y+ c_{1} \right )^{2}= c_{1}^{2}+ 4x\Leftrightarrow x= y\left ( y+ c_{2} \right )\Leftrightarrow 1= {y}'\left ( 2y+ c_{1} \right ),$ how can I break this cycle ?

答:
As the answers here have already given a hint of solving the equation in particular, I would like to clear the other doubt of how to break the cycle. But before that here is a complete solution of Albus Dumbledore 's answer-

I.F. = $e^{-\int \left(\frac{1}{y}\right)dy}=e^{-\ln y}=\frac{1}{y}$. And so the equation becomes $\frac{x}{y}=\int \left(\frac{y}{y}\right)dy$. That is $\frac{x}{y}=y+c$

We have got the original equation ${y}'\left ( x+ y^{2} \right )= y$-----------------------(1)

We also have got Wolfram alpha's solution $y'\left(2y+c\right)=1$--------------(2)

Now dividing (1)/(2).

We get $\frac{\left(x+y^2\right)}{\left(2y+c\right)}=y$

On rearranging we get  $\frac{\left(x+y^2\right)}{y}=\left(2y+c\right)$

Which on simplfying we get our solution

$\left(\frac{x}{y}\right)=\left(y+c\right)$


问:
Problem from Roman: a lower bound for trace of |T|^2
I'm working on a problem from Stephen Roman's Linear Algebra text, #20 on p. 235: Suppose $\tau \in \mathcal{L}(\mathbb{C}^n)$ and let the characteristic polynomial $\chi_{\tau}(x)$ have roots $\lambda_1, \dotsc, \lambda_n$ (with multiplicity). Show that $$\sum_{i=1}^n|\lambda_i|^2\leq \text{Tr}(\tau^*\tau),$$with equality holding iff $\tau$ is normal.

The last part is clear, since if $\tau$ is normal, then there is an ON basis $B$ diagonalizing $\tau$ and then $[\tau^* \tau]_B$ is diagonal with entries $|\lambda_i|^2$.

I'm trying to solve the first part. I note that $\text{Tr}(\tau^*\tau)$ is $\sum\limits_{i,j}|a_{ij}|^2$, if $(a_{ij})=[\tau]_B$ for $B$ an ON basis. I am also trying to exploit that $\tau^*\tau\geq0$. I'm stuck, though. Any advice?

答: Hint: Matrices over $\mathbb{C}$ can be unitarily triangularised. So, you may assume that the matrix of $\tau$ is lower triangular.</p>


问:
Composition of uniformly continuous functions
Composition of uniformly continuous functions.

I have seen many posts that seem to suggest that the composition of uniformly continuous functions is necessarily uniformly continuous(on R). However I'm pretty sure this is wrong and I'm hoping we can get a concrete answer. I think I have a counter example. Considering the set $[0,\infty)$, let $f(x)=e^{-x}$ and $g(x)=-x$ then $fg(x)=e^x$ is not uniformly continuous but $f$ and $g$ are.

答: That's not a counter example, since $g\bigl([0,\infty)\bigr)=(-\infty,0]$ and $f$ is not uniformly continuous on $(-\infty,0]$.


问:
Calculate the sum function of power series
I have the following series:
$$
\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}
$$
and I tried to calculate its sum function like this:
$$
\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}=\int\cdots\int\frac{d^n}{dx^n}\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}(dx)^n
$$
just by first taking nth derivative and then integrating n times. From what I've learnt, in the region of convergence of this power series (which is obviously $(-\infty, \infty)$), the following equation holds:
$$
\int\cdots\int\frac{d^n}{dx^n}\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}(dx)^n=\int\cdots\int\sum_{n=0}^{\infty}\frac{d^n}{dx^n}\frac{x^{2n}}{(2n)!}(dx)^n=\int\cdots\int\sum_{n=0}^{\infty}\frac{x^{n}}{n!}(dx)^n
$$
that is, you can change the order of taking derivative and summing, and $\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$ is just $e^x$, so integrate it n times should also be $e^x$. So the sum function (the way I see it) is just $e^x$.

But this is obviously wrong, since there is no way $\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}=e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$, but I don't know where is the mistake.

答:
There are two different $n$'s here:

the one from taking the $n$th derivative;

the one from the sum $\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}$.

Note that this last series is just a short way of writing$$1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots$$and that, if you differentiate this expression $n$ times you will not get $e^x$.


问:
An endomorphism of a field is the identity on a subfield. Is it an automorphism?
Let $K$ be a subfield of a field $L$, both infinite. Let $\phi$ be an endomorphism on $L$ such that $\phi|_{K} = id|_{K}$

Is it true that $\phi$ is an automorphism?

I believe, it is so if $L$ is a finite extension of $K$.
A counter example for finite $K$ and infinite $L$ would be the Frobenius's endomorphism on $\mathbb{Z}_2(t)$, which is identity on $\mathbb{Z}_2$, but not surjective.

答:
Your idea is right. Any endomorphism of a field $L$ induces the identity on the prime subfield, but need not be an automorphism if $L$ is infinite and your example is fine.

Indeed, if $L$ is a finite extension of $K$, then counting dimensions will show that an endomorphism inducing an automorphism on $K$ is necessarily surjective, because $[L:K]=[L:\phi(L)][\phi(L):\phi(K)]$ and $[\phi(L):\phi(K)]=[L:K]$.


问:
If a series is absolutely converge then the series can be regroup with changing their order?
I am just thinking about why this is true. Can I change it to

Q1. If a series is convergent then the series can be regrouped without changing the order of terms.

For example the sum of $(-1)^n$ is an alternating sequence and it is divergent, so I can't regroup them?

Q2. Can I claim that a convergent, non-alternating series be absolutely convergent?

As there is no difference after the term become absolute value, it should be still convergent after absolute those terms?

Q3. What does conditionally convergent mean?
If a sequence is either convergent or absolutely convergent then it is conditionally convergent?

答:
Answers:

An absolutely convergent series can be summed in any fashion. The value of the sum is not affected by order. Your supposition about $(-1)^n$ is completely correct. Based on regrouping and rearranging, you could assign a wealth of values to the series with $(-1)^n$ as the summand. Even a conditionally convergent series can be rearranged to give any value.

A non-alternating, convergent series is also absolutely convergent since you can either factor out a negative (if the series is made of only negative terms) or leave it as is and it will be absolutely convergent. It does not follow, however, that any convergent series is also absolutely convergent. Case in point: $\sum_{n=1}^{\infty} \frac{(-1)^n}{n} = \log(2)$, but the harmonic series diverges.

Conditionally convergent means that a series converges as is, but does not converge absolutely. Absolute convergence is much stronger than conditional convergence.


答:
An idea using a little basic complex analysis:

Clearly the convergence radius of this series is infinity, and

$$f(x)=\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}\implies f(ix)=\sum_{n=0}^\infty\frac{i^{2n}x^{2n}}{(2n)!}=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}=:\cos x$$

and the above is true in fact for any $\;x\in\Bbb C\;$ . Thus, we  get

$$f(ix)=\cos x=\frac{e^{ix}+e^{-ix}}2\implies f(x)=f((-i)ix)=\frac{e^{x}+e^{-x}}2=:\cosh x$$

Title: Calculate the sum function of power series

Body:
I have the following series:
$$
\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}
$$
and I tried to calculate its sum function like this:
$$
\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}=\int\cdots\int\frac{d^n}{dx^n}\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}(dx)^n
$$
just by first taking nth derivative and then integrating n times. From what I've learnt, in the region of convergence of this power series (which is obviously $(-\infty, \infty)$), the following equation holds:
$$
\int\cdots\int\frac{d^n}{dx^n}\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}(dx)^n=\int\cdots\int\sum_{n=0}^{\infty}\frac{d^n}{dx^n}\frac{x^{2n}}{(2n)!}(dx)^n=\int\cdots\int\sum_{n=0}^{\infty}\frac{x^{n}}{n!}(dx)^n
$$
that is, you can change the order of taking derivative and summing, and $\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$ is just $e^x$, so integrate it n times should also be $e^x$. So the sum function (the way I see it) is just $e^x$.

But this is obviously wrong, since there is no way $\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}=e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$, but I don't know where is the mistake.


问:
Find the differential equation in partial derivatives that follows from: $f(x+y+z)=xyz$
Find the differential equation in partial derivatives that follows from:

$$f(x+y+z)=xyz$$

Idea: If $u(x,y,z)=x+y+z$ and $v(x,y,z)=xyz$ then

$$\phi(u,v):=f(u)-v=0,$$ hence,

$$\frac{\partial\phi}{\partial x}=\frac{\partial \phi}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial \phi}{\partial v}\frac{\partial v}{\partial x}=f^\prime(u)+yz\frac{\partial \phi}{\partial v}$$ $$\frac{\partial\phi}{\partial y}=\frac{\partial \phi}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial \phi}{\partial v}\frac{\partial v}{\partial y}=f^\prime(u)+xz\frac{\partial \phi}{\partial v}$$

I think this is not the way to solve the problem, can someone help me please?

答:
$$f(x+y+z)=xyz\quad\text{or}\quad g(xyz)=x+y+x$$
with $f$ and $g$ inverse functions one to the other.

We are looking for a first order linear PDE on the general form :
$$\alpha(x,y,z)\frac{\partial F(x,y,z)}{\partial x}+\beta(x,y,z)\frac{\partial F(x,y,z)}{\partial y}+\gamma(x,y,z)\frac{\partial F(x,y,z)}{\partial z}=0$$
In order to simplify the editing we will use the notations :
$$\alpha F_x+\beta F_y+\gamma F_z=0$$

Moreover we look for $\alpha(x,y,z),\beta(x,y,z),\gamma(x,y,z)$ so that the above functions
$$u=x+y+z \text{ and } v=xyz$$
be the equations of characteristic curves so that the relationship $f(u)=v$ be satisfied.

If so, the general solution of the PDE is on the form of implicit equation :

$$F(x,y,z)=\Phi\big(u,v\big)=\Phi\big((x+y+z),(x,y,z)\big)$$

We compute the partial derivatives :
$$F_x=\Phi_u+yz\Phi_v$$
$$F_y=\Phi_u+xz\Phi_v$$
$$F_z=\Phi_u+xy\Phi_v$$
From the three equations one can eliminate $\Phi_u$ and $\phi_v$ which leads to the next relationship :
$$\boxed{x(z-y)F_x+y(x-z)F_y+z(y-x)F_z=0}$$
So, the coefficients of the PDE are determined :
$$\alpha=x(z-y)$$
$$\beta=y(x-z)$$
$$\gamma=z(y-x)$$


答:
Following up on the previous answer for the sake of those perhaps looking how to actually find what is the function in the series without complex analysis. Note that if
$$
f(x)
 = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}
 = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \ldots,
$$
which is just the listing of all even powers of
$$
e^x
 = \sum_{n=0}^\infty \frac{x^n}{n!}
 = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots
$$
The trick to make $f(x)$ from $e^x$ is to find a way to negate the odd terms, which is to multiply all of them by $-1 = (-1)^n$ for odd $n$. Conventiently, for even $n$, we have $(-1)^n=1$ so there will be no change to even terms:
$$
e^{-x}
 = \sum_{n=0}^\infty \frac{(-x)^n}{n!}
 = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} \pm \ldots
$$
Now, adding $e^x$ and $e^{-x}$ doubles the even terms, and cancels the odd terms. To "undo" the doubling, we halve the result to get
$$
\frac{e^x+e^{-x}}{2}
 = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \ldots,
 = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}
 = f(x),
$$
so $f(x) = \cosh x$.


问:
Finding $\operatorname{Im}T$ - Linear Algebra
Regarding :

$T: \mathbb{R}^{n+1} \to \mathbb{R}^{n}, T(x_1,...,x_{n+1})=(x_1+x_{n+1},x_2,...,x_n) $

I need to find $\ker T$ and $\operatorname{Im}T.$

To find $\operatorname{Im}T,$ I want to take a spanning set of $\mathbb{R}^{n+1}$, which will be $(e_1,e_2,...,e_n,e_{n+1}),$ and by a theorem. I can say that :

$\operatorname{Im}T=\operatorname{span}\{T(e_1),...,T(e_{n+1})\}=\operatorname{span}\{e_1+e_{n+1},e_2\ldots, e_{n}\}$

But now I have a problem, because those vectors doesn't belong to $\mathbb{R}^n.$

Any help ?

Thanks a lot!

答:
You have to distinguish between the vectors in a basis of $\mathbb{R}^{n+1}$ and those in $\mathbb{R}^n$. If you “prime” the ones in the codomain, you have
$$
T(e_1)=e_1',\quad T(e_2)=e_2',\quad \dots,\quad T(e_n)=e_n',\quad T(e_{n+1})=e_1'
$$

Thus you see that $\{e_1',e_2',\dots,e_n'\}$ is contained in the image, so the linear map is surjective. Thus, by rank-nullity, its kernel has dimension $1$ and it suffices to find a nonzero vector in the kernel, which is, for instance, $e_1-e_{n+1}$, so the kernel is generated by $e_1-e_{n+1}$.


问:
Every point of a grid is colored in blue, red or green. How to prove there is a monochromatic rectangle?
I have a $3$-coloring of $\mathbb{Z}\times\mathbb{Z}$, i.e. a function $f:\mathbb{Z}\times\mathbb{Z}\to\{\color{red}{\text{red}},\color{green}{\text{green}},\color{blue}{\text{blue}}\}$.

I have to prove that there is a monochromatic rectangle with its sides being parallel to the axis, i.e. to prove that for some choice of $a,b,c,d\in\mathbb{Z}$ with $a\neq b$ and $c\neq d$, all the points $(a,c),(a,d),(b,c),(b,d)\,$ have the same color.

I tried to work by contradiction, without achieving much.

Additionally, can we prove some upper bound on $|a-b|$ and $|c-d|$?

答:
We will decide how to choose rows first. So basically there are three colors so applying the pigeonhole principle if we choose the minimum number of rows to be 4 then it is bound at the worst case one color will repeat.

Now for choosing the minimum number of columns out of 4 rows we will choose any 2 rows and assign its color to be the same, either red blue, or green. This will lead us to the combination of 18 possible options (4C2*3 (for each color)).

So again applying pigeonhole there should be at least 19 (18+1) columns for the possibility of two colors in 4 rows to be the same


问:
Can't seem to solve $e^{3e^{\frac{i\pi}{3}}}$ into cartesian form
I tried to use the code to format the equations correctly but I can't seem to make it work so I'm just going to type normally.
So I have a question on $e^{3e^{\frac{i\pi}{3}}}$. I was going to do it the normal way such that it equals to $\cos(3e^{\frac{i\pi}{3}})+i\sin(3e^{\frac{i\pi}{3}})$, but its not working since I can't exponentially multiply it. Are there any other methods that I can use?

Body:
Note that
$$
3e^{\frac{\pi }{3}i}  = \frac{3}{2} + \frac{{3\sqrt 3 }}{2}i.
$$
Thus,
$$
e^{3e^{\frac{\pi }{3}i} }  = e^{\frac{3}{2}} e^{\frac{{3\sqrt 3 }}{2}i}  = e^{\frac{3}{2}} \cos \left( {\frac{{3\sqrt 3 }}{2}} \right) + ie^{\frac{3}{2}} \sin \left( {\frac{{3\sqrt 3 }}{2}} \right).
$$


问:
If $f \in L((0,1)^n)$ bounded does follow that $f$ is in $L(\mathbb{R}^n)$?
I read a few things about $L^2$-Spaces and I am not at all sure whether I understand it right.
So here are two question which I struggle with:

If a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is positive and bounded, is $f\in L^2((0,1)^n)$?

And are bounded functions in $ L^2((0,1)^n)$ also in $L^2(\mathbb{R}^n)$?

答: For your first question : yes because constant function are $L^2((0,1)^d)$. For your second, no. Take for example $f(x)=x^2$ (with $d=1$).


问:
Can I perform topological sorting on cycle graph?
I am told to perform a attempted topological sorting on cycle graph (non acyclic). Can I do so? Since a topological sorting only can be perform on a dag(directed acyclic graph).

Click here to view the question with the graph

答:
I have created a cyclic-toposort project that aims to sort directed cyclic graphs by determining the least amount of cyclic edges that would result in a sortable directed acyclic graph. I published the cyclic_toposort project on github. It not only quickly sorts a directed cyclic graph providing the minimal amount of violating edges but also optionally provides the maximum groupings of nodes that are on the same topological level (and can therefore be activated at the same time) if desired.

If the problem is still relevant to you then I would be happy if you check out my project and let me know what you think!

This project was useful to my own neural network topology research, so I had to create something like this anyway. I am answering your question this late in case anyone else stumbles upon this thread in search for the same question.


问:
Calculating $\Vert\sum_n^\infty(-1)^nx_n \Vert_\infty$
I want to calculate norm of functional:

$$f:(l_1,\Vert\cdot\Vert_\infty) \ni x_n\rightarrow \sum_n^\infty(-1)^nx_n $$

My work so far

Let's firstly check if this funcional is well defined. To do so let's consider

$$\sum_n^\infty |(-1)^nx_n|= \sum_n^\infty|x_n| <\infty\;\;\; \text{because we are in $l_1$}$$

Let's now consider $|f(x_n)|$ (to calculate norm of functional we want to find $\sup_{\{x_n:\sup_n|x_n| \le 1\}}|f(x_n)|$):

$$|f(x_n)| = |x_1+x_2+x_3+...| \le |x_1| + |x_2|+... \le \Vert x_n \Vert_\infty + \Vert x_n \Vert_\infty +...$$

Now we want to consider such sequences for which $\sum_n^\infty|x_n|  < \infty$ and $\sup_n|x_n| \le 1$.

My question is: how can i choose such sequence which satisfies conditions above as well as  $|f(x_n)| =  \Vert x_n\Vert_\infty + \Vert x_n\Vert_\infty+...$

Would it be sufficient to calculate $\Vert f \Vert$ ?

答: Call $x^m_n=\begin{cases}0&\text{if }n>m\lor 2\nmid n\\ 1&\text{if }n\le m\land 2\mid n\end{cases}$. Then it is clear that $\lVert x^m\rVert_\infty\le1$ and $f(x^m)=\sum_{n=1}^\infty (-1)^nx_n^m=\left\lfloor \frac m2\right\rfloor$, and therefore $$\sup_{x\in l_1,\ \lVert x\rVert_\infty\le 1}\lvert f(x)\rvert\ge\sup_{m\in\Bbb N}\lvert f(x^m)\rvert=\infty$$


答: This is not a continuous linear functional, so its  norm is $\infty$. If there is a constant $C$ such that $|\sum (-1)^{n}x_n| \leq C\sup |x_n|$ for all $(x_n) \in \ell^{1}$ the we can take $x_n=(-1)^{n}$ for $n \leq N$ and $x_n=0$ for $n >N$ to see that $N\leq C$. This cannot hold for all $N$ so $f$ is not continuous.

Title: Calculating $\Vert\sum_n^\infty(-1)^nx_n \Vert_\infty$

Body:
I want to calculate norm of functional:

$$f:(l_1,\Vert\cdot\Vert_\infty) \ni x_n\rightarrow \sum_n^\infty(-1)^nx_n $$

My work so far

Let's firstly check if this funcional is well defined. To do so let's consider

$$\sum_n^\infty |(-1)^nx_n|= \sum_n^\infty|x_n| <\infty\;\;\; \text{because we are in $l_1$}$$

Let's now consider $|f(x_n)|$ (to calculate norm of functional we want to find $\sup_{\{x_n:\sup_n|x_n| \le 1\}}|f(x_n)|$):

$$|f(x_n)| = |x_1+x_2+x_3+...| \le |x_1| + |x_2|+... \le \Vert x_n \Vert_\infty + \Vert x_n \Vert_\infty +...$$

Now we want to consider such sequences for which $\sum_n^\infty|x_n|  < \infty$ and $\sup_n|x_n| \le 1$.

My question is: how can i choose such sequence which satisfies conditions above as well as  $|f(x_n)| =  \Vert x_n\Vert_\infty + \Vert x_n\Vert_\infty+...$

Would it be sufficient to calculate $\Vert f \Vert$ ?


问:
Ratio of two line segments in a special right triangle
So I had this question in my mind for a while.

At first, I tried using a bunch of theorems such as the Law of Sines, the Law of Cosines, and similarity. I got the answer $\frac{a}{a-1}$, but then I realized it was wrong because of a factoring issue.

答:
The triangle $\,ABC\,$ is similar to
$\,EDC.\,$ Use similarity of sides to get
$$ \frac{r_1}{a+r_1} = \frac{r_2}a = \sin(\alpha). \tag{1} $$
This implies
$$ 1-\sin(\alpha) = 1 - \frac{r_2}a = \frac{a-r_2}a. \tag{2} $$
Equality of ratios in equation $(1)$ gives us
$$ r_1\,a= (a+r_1)\,r_2 = a\,r_2 + r_1r_2. \tag{3} $$
Rearrange terms and and add $\,a^2\,$ to get
$$ a^2 = a^2 + r_1\,a - a\,r_2 - r_1r_2 = (a+r_1)(a-r_2). \tag{4} $$
This implies the equality of ratios
$$ \frac{a+r_1}a = \frac{a}{a-r_2}. \tag{5} $$
Combine equality of ratios in equation $(1)$ with equations $(2)$ and $(5)$ to get
$$ \frac{r_1}{r_2} = \frac{a+r_1}a= \frac{a}{a-r_2} = \frac1{1-\sin(\alpha)}. \tag{6} $$


问:
Atomic measure on generators
In measure theory, I've sometimes seen arguments of the sort of the following, if I'm interpreting it correctly.

Let $(X,\Sigma)$ be a measurable space where the $\sigma$-algebra $\Sigma$ is countably generated. Let $\mu$ be a probability measure on $\Sigma$, and suppose that for a set of generators $\{S_i\}_{i\in\mathbb{N}}$ of $\Sigma$ we have that for all $i\in\mathbb{N}$, $\mu(S_i)\in\{0,1\}$.
Then for all measurable sets $A\in\Sigma$, $\mu(A)\in\{0,1\}$ as well.

First of all, is this correct? If so:

How can we prove it?

Is the "countably generated" assumption necessary?

答: Same as "always" when you're proving things about the $\sigma$-algebra generated by some family of sets: Let $\mathcal A=\{E\in\Sigma:\mu(E)\in\{0,1\}\}$. Show that $\mathcal A$ is a $\sigma$-algebra. Now since $$\{S_i\}\subset\mathcal A\subset\Sigma,$$the definition of "$\Sigma$ is generated by $\{S_i\}$" says that $\mathcal A=\Sigma$.

Title: Atomic measure on generators


问:
What is correct definition of moment generating function.
I had searched the literature and found two different definition of Moment Generating Function (MGF).

$$\phi_{w}(-s)=\int_0^{\infty}\text{exp}(-sw)f(w)\text{d}w \tag{1} $$

$$\phi_{w}(s)=\int_0^{\infty}\text{exp}(sw)f(w)\text{d}w \tag{2} $$

Out of these, which one is correct definition. Any help in this regard is highly appreciated.

答:
both are wrong. The correct definition is the following

$$M_X(t)=\mathbb{E}[e^{tX}]$$

when the expectation exists... not necessarily the expectation is an integral

for example, let $X\sim B(1;\theta)$, that is a bernoulli with parameter $\theta$, its MGF is given by

$$M_X(t)=e^0\cdot (1-\theta)+ e^t\cdot \theta=(1-\theta)+e^t\cdot \theta$$


问:
Suppose $F$ is a field with $p^n$ elements where $p$ is prime. Let $V$ be a $k$-dimensional vector space over $F$.
Suppose $F$ is a finite field with $p^n$ elements where $p$ is prime. Let $V$ be a $k$-dimensional vector space over $F$. Then what is the cardinality of the following?

The number of linear transformations $T:V \rightarrow V$

The number of invertible linear transformations $T:V \rightarrow V$

The number of linear transformations $T:V \rightarrow V$ with determinant 1.

I have tried to solve it by trying to count the number of matrix representations (of linear transformations), but even if I manage to do that, a linear transformation can have more than one matrix representation.

答:
Hint

A linear map is fully and uniquely defined by the image of the canonical basis $\{e_1, \dots, e_k\}$. And there are $(p^n)^k = p^{nk}$ vectors in $V$.

Point 1

Each $e_i$ may have $ p^{nk}$ images. Therefore there are $p^{nk^2}$ linear transformations.

Point 2

The image of $e_1$ can be every vector except the zero one. The image of $e_2$ can be what you want providing that it doesn't lie in the image of $e_1$. Similarly for the other vectors.

Point 3

The determinant is a group homomorphism from $\mathcal L(V,V)$ to $F$. Its kernel if $\{1\}$.


问:
Conic Sections formulae using Complex Numbers
I have been preparing for the JEE Examination here in India and have been studying Complex Numbers for the past few days. One of the topics which falls under Complex Numbers is their application in Coordinate Geometry (Conic Sections). These include the following:

Equation of lines in different forms (parametric/non-parametric)$$$$
    Equations of Circles with various conditions (for example centered at $z_0$, or orthogonal to another circle and so on)$$$$
    Equations of Ellipses and Hyperbolas and so on.

$$$$
There are several other applications/equations mentioned which I haven't written here. The trouble I'm facing is that the book I use for Complex Numbers (Algebra for JEE Main and Advanced, by SK Goyal) only has the formulae listed without showing the derivations. I find it extremely hard to just accept and use a formula/result without knowing how it came into existence.

$$$$

I would be grateful if somebody could please mention a source/reference book from which I could actually learn how to derive/reach all the formulae used for represinting Conics using Complex Numbers. The book does not necessarily have to match the level of the JEE Advanced Examination; it can be higher than that too. However I would prefer it if the book was at the level suitable for the JEE Advanced level only.

$$$$Many thanks in advance!
$$$$
Edit:$$$$ The results mentioned in my book are as follows:
$$$$

$1) $The equation of the line joining $z_1$ and $z_2$ is $$z(\bar{z_1}-\bar{z_2})-\bar{z}(z_1-z_2)+ z_1\bar z_2-z_2\bar z_1=0 \text{ (non parametric form).}$$

$$$$

$2) $Three points are collinear if $$\begin{vmatrix}z_1&\bar{z_1}&1\\z_2&\bar {z_2}& 1\\z_3&\bar {z_3}& 1\end{vmatrix}=0$$

$$$$

$3)$$\bar{a}z+\bar z a+b=0$ where $b\in \mathbb R$ describes the equation of a straight line (I don't know what $a$ is, nor what $\iota b$ is).

$$$$

$4)$ The real and complex slope (I don't know what is meant by 'real' and 'complex') of the line $\bar{a}z+\bar z a+b=0$  are $-\dfrac{\Re(a)}{\Im(a)}$ and $-\dfrac{a}{\bar a}$ where $b\in \mathbb R$.

$$$$

$5) $If the lines $\bar{a}z+\bar z a+k_1=0$ and $\bar{b}z+\bar b a+k_2=0$ $k_1,k_2\in \mathbb R$ are perpendicular to each other, then $$\bar{a}b+\bar ba=0$$

$$$$

$6)$ $z\bar z +a\bar z +\bar az+k=0 $ where $k\in \mathbb R$ represents a circle with center $-a$ and radius $\sqrt{|a|^2-k}$.

$$$$

$7)$ If $|z-z_1|+|z-z_2|=2a$ where $2a>|z_1-z_2|$ then $ z $ represents an ellipse with foci at $z_1 \text{ and }z_2$ and $a\in \mathbb R^+ $.

$$$$

$8)$ If  $|z-z_1|-|z-z_2|=2a$ where $2a<|z_1-z_2|$ then $z$ represents a hyperbola with foci at $z_1$ and $z_2$.

$$$$

$9)$ Equation of all circles orthogonal to $|z-z_1|=r_1\text{ and }|z-z_2|=r_2$ is (nothing further is mentioned).

$$$$

$10)$ $\left |\dfrac{z-z_1}{z-z_2}\right | = k$ is a circle if $k\neq 1$ and will represent a line if $k=1$.

$$$$

$11)$ The equation $|z-z_1|^2+|z-z_2|^2=k$ will represent a circle if $k\geq \frac12 |z_1-z_2|^2$.

$$$$

$12)$ If $\arg\left(\dfrac{(z_2-z_3)(z_1-z_4)}{(z_1-z_3)(z_2-z_4)}\right)=0, \pm \pi$, then the points $z_1,z_2,z_3,z_4$ are concyclic.

答:
There's detail discussion for general conics in Masayo Fujimuraa's paper:

The Complex Geometry of Blaschke Products of Degree 3 and Associated Ellipses


问:
Equality of expressions containing supremum of some double sums
I am doing a project related to operator norm and some double sequences. In the course of proving some results, I encounter the following expressions:

$\displaystyle \|\alpha\| := \sup_{\|x\|_p=1} \left[\sum_{j=1}^\infty \left|\sum_{i=1}^\infty x_i \alpha_{ij}\right|^q\right]^{\frac{1}{q}}$

and

$\displaystyle \|\alpha\|^* := \left(\sum_{j=1}^\infty \sum_{i=1}^\infty |\alpha_{ij}|^q \right)^{\frac{1}{q}}$,

where $\|\cdot\|_p$ is the $\ell_p$ norm, $p$ is the conjugate of $q$, and $x=(x_i)$ is a sequence, $\alpha=(\alpha_{ij})$ is a double-indexed sequence.

I want to say that they are equal. But is it true that $\|\alpha\|= \|\alpha\|^*$ ?

Thank you.

Body:
You can take $\alpha_{11}=\alpha_{22}=1$, and all other coefficients equal to $0$. Let us fix $p=q=2$. Then
$$
\|\alpha\|^*=2^{1/2},
$$
while, for $x$ with $\|x\|_2=1$,
$$
\left[\sum_{j=1}^\infty \left|\sum_{i=1}^\infty x_i \alpha_{ij}\right|^2\right]^{\frac{1}{2}}=(|x_1|^2+|x_2|^2)^{1/2}\leq\|x\|_2=1.
$$
So $\|\alpha\|=1$.


问:
With a perfect square containing one quadrant of a perfect circle, can it be easily determined whether a point is inside the circle?
Diagram: https://i.imgur.com/uOqjSk8.jpg

This is an easy problem if you compute the distance from the corner of the square at the center of the circle (just whether d is less-than-or-equal to r), but is it possible to determine from the point opposite that one without computing the distance from the circle's center.

The context is for image processing. I have the four outer corners of a bounding box but any algorithmic solution becomes complex when you need to compute which part of the bounding box a particular point is. However, I know that every point is inside the bounding box, so I'm wondering if it's possible to measure the distance from all four outer points and determine if a point would lie outside a circle based on its distance from those four corners.

Body:
For distances to two adjacent vertices:
$$
d_1^2=(r-x)^2+(r-y)^2,\qquad d_2^2=(r+x)^2+(r-y)^2
$$
so one can find that,
$$
2(x^2+y^2) = d_1^2 + d_2^2 - \sqrt{(2r+d_1+d_2)(d_1+d_2-2r)(2r+d_1-d_2)(2r+d_2-d_1)}
$$

From $2(x^2+y^2)>2r^2$, we find the critertion for the point to be out of the cirle:
$$
d_1^2 + d_2^2 - \sqrt{(2r+d_1+d_2)(d_1+d_2-2r)(2r+d_1-d_2)(2r+d_2-d_1)} > 2r^2
$$

Or if you simplify:
$$
d_1^4+d_2^4 +10r^4 > 6(d_1^2+d_2^2)r^2.
$$


问:
Conditional Expectation: $E(X|Y) \geq Y$ and $E(Y|X) \geq X$ implies $X=Y$ a.e.
Suppose for random variables $X$ and $Y$ with finite expectation we he have $E(X|Y) \geq Y$ and $E(Y|X) \geq X$.  We want to show that $X=Y$ almost everywhere.

My attempt:

Let $A \in \sigma(X)$ and $B \in \sigma(Y)$.  Then we have

$\int_B X = \int_B E(X | Y) \geq \int_B Y$ and $\int_A Y = \int_B E(Y | X) \geq \int_A X$.

My idea now is to consider sets of the form $\{X>a>Y\}$ for $a$ rational but I'm having trouble with getting the other variable involved.  The issue seems to be that $\{X>a\}$ is in $\sigma(X)$ but not necessarily in $\sigma(Y)$ and it seems like we cannot get the other inequality working.  Any help would be appreciated.

答:
Let $(\Omega, \Sigma, P)$ be a probability space.

Suppose for random variables $X$ and $Y$ with finite expectation we he have $E(X|Y) \geq Y$ and $E(Y|X) \geq X$.  We want to show that $X=Y$ almost everywhere.

For any  $A \in \sigma(X)$ and any $B \in \sigma(Y)$. We have

$\int_B X = \int_B E(X | Y) \geq \int_B Y$ and $\int_A Y = \int_A E(Y | X) \geq \int_A X$.

Since $X$ and $Y$ have finite expectation, we can conclude that, for any  $A \in \sigma(X)$ and any $B \in \sigma(Y)$,

$$\int_B (X -  Y)  = \int_B X - \int_B Y \geq 0$$ and $$\int_A (X - Y)= -\left( \int_A Y - \int_A X\right) \leq 0$$

Since $\Omega \in \sigma(X) \cap \sigma(Y)$, it follows immediately that $\int_\Omega (X -  Y) =0$.

Since for all  $B \in \sigma(Y)$,  $B^c \in \sigma(Y)$, we have
$\int_B (X -  Y) \geq 0$, $\int_{B^c} (X -  Y) \geq 0$ and
$$ \int_B (X -  Y)  + \int_{B^c} (X -  Y) = \int_\Omega (X -  Y) =0 $$

So for all $B \in \sigma(Y)$ we have
$$\int_B (X -  Y) = 0  \tag{1}$$

In a similar way, since for all  $A \in \sigma(X)$,  $A^c \in \sigma(X)$, we have
$\int_A (X -  Y) \leq 0$, $\int_{A^c} (X -  Y) \leq 0$ and
$$ \int_A (X -  Y)  + \int_{A^c} (X -  Y) = \int_\Omega (X -  Y) =0 $$

So for all $A \in \sigma(X)$ we have
$$\int_A (X -  Y) = 0 \tag{2} $$

From $(1)$ and $(2)$, we have that for all $C \in  \sigma(X) \cup  \sigma(Y)$,
$\int_C (X -  Y) = 0 $.

Now, note that $\Gamma = \{ D \in \Sigma : \int_D (X -  Y) = 0 \}$ is a $\sigma$-algebra and, since $\sigma(X) \cup  \sigma(Y) \subseteq \Gamma$, we have that $\sigma( \sigma(X) \cup  \sigma(Y)) \subseteq \Gamma$.

So, for all $C \in \sigma(X, Y) = \sigma( \sigma(X) \cup  \sigma(Y))$, we have
$\int_C (X -  Y) = 0 $.

Now, note that $[X \geq Y]= \{ w \in \Omega : X(w) \geq  Y(w)\} \in \sigma(X, Y)$ and
$[X < Y]= \{ w \in \Omega : X(w) <  Y(w)\} \in \sigma(X, Y)$. So
$$\int_\Omega |X-Y| = \int_{[X \geq Y]} (X-Y) + \int_{[X < Y]} (-(X-Y)) = \int_{[X \geq Y]} (X-Y) - \int_{[X < Y]} (X-Y)  =0$$

So, $X=Y$ almost everywhere.


问:
Why does Cantor's diagonal argument not work for rational numbers?
If we map every integer to a string that represents a rational number, and produce a number different from all the ones listed, we are essentially following Cantor's algorithm. But why does it not apply? Is it because we can't be certain that the number produced is a rational number?</p>

答:
It applies, in the sense that you can carry it out. However, the number you obtain through the process is not a rational (it does not have a periodic decimal expansion).

To be precise, the procedure does not let you guarantee that the number you obtain has a periodic decimal expansion (that is, that it is a rational number), and so you are unable to show that the "diagonal number" is a rational that was not in the original list. In fact, if your original list is given explicitly by some bijection, then one is able to show just as explicitly that the number you obtain is not a rational.


问:
$\log_x(y)=m^4$ and $\log_y(x)=5/(m^3)$, what is the value of $m$?
$\log_x(y)=m^4$ and $\log_y(x)=5/(m^3)$, what is the value of $m$? </p>

答:
stahl and response wrote almost full answer so I'll just add final steps:
$log_x(y)=m^4,log_y(x)=\frac 5 {m^3}\rightarrow m^4=log_x(y)=\frac {log_y y}{log_y x}=\frac 1 {\frac 5 {m^3}}=\frac {m^3} {5}$ and from here that's clear that $m^3=5m^4 \rightarrow m^3(5m-1)=0$ which means that the solutions are: $m=0,\frac 1 5$

EDIT:

as adriano noted, since $log_y(x)=\frac 5 {m^3}$ we reject $m=0$ so the final solution is $m=\frac 1 5$


问:
Limit as $x\to \infty,$ of $\;\frac{(1+\frac{1}{x})^{x}}{x}$
I'm having some trouble evaluating the $\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\right)^{x}}{x}$ in the context of real analysis. Denote this sequence by $(a_k)$

If so, then I need to prove that for all $\varepsilon>0$

$$\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right| <\varepsilon$$

We know that $\lim_{x\to\infty}1/x=0$, however $|1/x|\leq\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}\right|$, so I'm not sure this information is useful. I considered the sequential characterization of continuity, so if the sequence above converges to 0, then $f((a_k))$ converges to $f(0)$, but there is also not an obvious choice of $f$. Any suggestions into how to evaluate the limit?

Body:
$$y=\frac{\left(1+\frac{1}{x}\right)^x}{x}\implies \log(y)=x \log \left(1+\frac{1}{x}\right)-\log(x)$$ So, by Taylor
$$\log(y)=x\Bigg[\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}+O\left(\frac{1}{x^4}\right)\Bigg]-\log(x)$$
$$\log(y)=1-\log(x)-\frac{1}{2 x}+\frac{1}{3
   x^2}+O\left(\frac{1}{x^3}\right)$$
$$y=e^{\log(x)}=\frac{e}{x}\left(1-\frac{1}{2 x}+\frac{11}{24 x^2}+O\left(\frac{1}{x^3}\right) \right)\sim \frac{e}{x} \quad \to 0$$


问:
How to break the components of a vector perpendicular to another?

Here , how should we break down the components of T here. Like we can’t make a triangle here since angle of mg with T here is 90. Then , how should we draw and make the T components.

2nd diagram just tells us that if there was an angle theta. Then , it was easy to Ames the components.

答: Generally, you don't break a vector up into components if it is already lined up with an axis.  If you have $m\overrightarrow{g}$ pointing in the $-y$ direction, for example, and $\overrightarrow{T}$ points in the $+x$ direction, then there's no need to break them up.  The components of $m\overrightarrow{g}$ in that case would be $(0, -mg)$ and the components of $\overrightarrow{T}$ would be $(T,0)$.


问:
How can I find the surface area of a normal chicken egg?
This morning, I had eggs for breakfast, and I was looking at the pieces of broken shells and thought "What is the surface area of this egg?" The problem is that I have no real idea about how to find the surface area.

I have learned formulas for circles, and I know the equation for an ellipse; however, I don't know how to apply that.

The only idea I can think of is to put an egg on a sheet of paper and trace it, and then measure the outline drawn, and then try to find an equation for that ellipse and rotate that about the $x$-axis.  Now, my problem is how I can find the equation of the ellipse from the graph, and will my tracing method really be the edge of the egg?  Also, can I use the standard surface area integral? Will I have to use some techniques to solve the integral that are not covered in the AP BC Calculus?

There has to be a better method for finding the surface area. Please, help me understand how to find the surface area of an egg; i.e., how to use my mathematical knowledge for something other than passing exams.

答:
Here's a shot:

Place dots on the egg, fairly close together, and connect them to form "triangles" on the egg. Measure the distances between the points of the points in the triangle and figure out the area as if it were a Euclidean triangle. Do this for several eggs, perhaps dividing by the square of the length along an axis to normalize it. Then repeat the process, with smaller and smaller triangles, until (hopefully!) the numbers will sort of agree up to some decimal place of accuracy.


问:
If $f$ is a polynomial of degree $k$, can we quickly compute $f(1),f(2),\dots,f(n).$
$f(x)=\left(\sum_{i=1}^k{a_ix^i}\right) \pmod P$

$P > n \gg k.$

Obviously calculating $f(1),f(2)..f(n)$ takes $O(nk)$.

So is there a faster way to speed up the process? $k$ is not very big. $k$ is only about $30.$

答:
Thanks for your reply.In fact, both n and P are large, P>n, k is small.

I know how to use FFT to speed up multipoint evaluations of polynomials, it takes $O(nlog^2n)$.

Because k is small here, and x is 1,2,3...,n, so I guess there's a faster way to do it.


问:
Probability of ending with 2 blue balls in a row after $n$ draws
Thank you for taking the time to read my question.

The scenario is, we're playing a game where we have in a box 9 balls: 4 red, 3 green, 2 blue.

We draw one ball at a time, and note down its color and then return it for the next draw, the game stops only when we have drawn the same color twice in a row.

What is the probability of the game ending after $n$ draws? $P(X=n)$?

What is the probability of the last two balls being blue?

I have tried to treat it as a geometric probability distribution, with: p: probability of having two balls in a row with the same color $P(X=n) = n-1$ fails and then a success but this ball number 2 and 3 could be the same color in this. I've tried to discuss it case by case, starting with if the first ball is Blue, Red, Green, in hopes of noticing some recursion, but no luck. Haven't had any other ideas.

答:
You can treat the process as a $6$-state Markov chain.  If $\ B_t\ $ is the colour of the ball drawn at time $\ t\ $, the states are
\begin{align}
\text{State $1$:}\ &B_t=\text{red}\ \ \text{ and either }\ \ t=0\ \ \text{ or }\ \ B_{t-1}\ne\text{red}\\
\text{State $2$:}\ &B_t=\text{green}\ \ \text{ and either }\ \ t=0\ \ \text{ or }\ \ B_{t-1}\ne\text{green}\\
\text{State $3$:}\ &B_t=\text{blue}\ \ \text{ and either }\ \ t=0\ \ \text{ or }\ \ B_{t-1}\ne\text{blue}\\
\text{State $4$:}\ &B_t=B_{t-1}=\text{red}\\
\text{State $5$:}\ &B_t=B_{t-1}=\text{green}\\
\text{State $6$:}\ &B_t=B_{t-1}=\text{blue}\\
\end{align}
States $4$, $5$ and $6$ are absorbing states—where the game ends—, the transition matrix is
$$
P=\pmatrix{0&\frac{1}{3}&\frac{2}{9}&\frac{4}{9}&0&0\\
          \frac{4}{9}&0&\frac{2}{9}&0&\frac{1}{3}&0\\
        \frac{4}{9}&\frac{1}{3}&0&0&0&\frac{2}{9}\\
    0&0&0&1&0&0\\
    0&0&0&0&1&0\\
    0&0&0&0&0&1}\ ,
$$
and the initial state probabilities are
$$
\pi_0=\pmatrix{\frac{4}{9}&\frac{1}{3}&\frac{2}{9}&0&0&0}\ .
$$
The state probabilities $\ \pi_n\ $ after the $\ (n+1)^\text{th}\ $ draw are given by
$$
\pi_n=\pi_0P^n\ ,
$$
and the probability that the game ends on or before the $\ (n+1)^\text{th}\ $ draw (for $\ n\ge1\ $) is
$$
\pi_{n4}+\pi_{n5}+\pi_{n6}\ .
$$
The probability $\ f_n\ $ that it ends precisely on the $\ (n+1)^\text{th}\ $ draw is therefore given by
$$
f_n=\pi_{n4}+\pi_{n5}+\pi_{n6}-\left(\pi_{n-1\,4}+\pi_{n-1\,5}+\pi_{n-1\,6}\right)\ .
$$
Magma gives $$\ x^4-x^3-\frac{26}{81}x^2+\frac{62}{243}x+\frac{16}{243}=
(x-1)\left(x^3-\frac{26}{81}x-\frac{16}{243}\right)
$$
as the minimal polynomial of $\ P\ $. This means that $\ f_n\ $ will satisfy the recursion
$$
f_{n+4}=f_{n+3}+\frac{26}{81}f_{n+2}-\frac{62}{243}f_{n+1}-\frac{16}{243}f_n\ ,
$$
so another way of calculating $\ f_n\ $ is to use the equation
$$
\pmatrix{f_{n+1}\\f_n\\f_{n-1}\\f_{n-2}}=\pmatrix{0&1&0&0\\
             0&0&1&0\\
             0&0&0&1\\
             1&\frac{26}{81}&-\frac{62}{243}&-\frac{16}{243}}^n\pmatrix{f_n\\f_{n-1}\\f_{n-2}\\f_{n-3}}\ ,
$$
with initial condition
$$
\pmatrix{f_1\\f_2\\f_3\\f_4}=\pmatrix{\frac{29}{81}\\
\frac{2}{9}\\
\frac{970}{6561}\\
\frac{1868}{19683}}\ .
$$
An alternative is to express $\ f_n\ $ as a linear combination of powers of the roots $\ 1,\lambda_1, \lambda_2, \lambda_3\ $ of the minimal polynomial:
$$
f_n=A+B\lambda_1^n+C\lambda_2^n+D\lambda_3^n\ .
$$
Magma gives $\ 0.64985, -0.39017$ and $\ -0.25969\ $ as the approximate values of $\ \lambda_1, \lambda_2$ and $\ \lambda_3\ $ rounded to $5$ decimal places.  Since the absolute values of these are all strictly less than one, then $\ \lim_\limits{n\rightarrow\infty}f_n=A\ $, and since we know this limit must be $0$, then we have $\ A=0\ $. But, in any case, we can determine the values of $\ A,B,C$, and $\ D\ $, from the initial conditions:
$$
\pmatrix{1&\lambda_1^2&\lambda_2^2&\lambda_3^2\\
         1&\lambda_1^3&\lambda_2^3&\lambda_3^3\\
         1&\lambda_1^4&\lambda_2^4&\lambda_3^4\\
         1&\lambda_1^5&\lambda_2^5&\lambda_3^5}\pmatrix{A\\B\\C\\D}=\pmatrix{\frac{29}{81}\\
\frac{2}{9}\\
\frac{970}{6561}\\
\frac{1868}{19683}}\ .
$$
Magma's approximate solution to these equations rounded to $5$ decimal places is $\ A=0, B=0.82440, C=0.03486\ $, and $\ D=0.08024\ $.

You can find the probability of the last two balls being blue (i.e. that the final state is state $6$) by solving a set of linear equations for the conditional probabilities $\ s_i\ $ that the final state is state $6$ given that the initial state is state $\ i\ $ for $\ i=1,2,3\ $ and then computing $\ \frac{4s_1}{9}+\frac{s_2}{3}+\frac{2s_3}{9}\ $. The equations for $\ s_i\ $ are
\begin{align}
s_1&=\frac{s_2}{3}+\frac{2s_3}{9}\\
s_2&=\frac{4s_3}{9}+\frac{2s_3}{9}\\
s_3&=\frac{4s_3}{9}+\frac{s_2}{3}+\frac{2}{9}\ ,
\end{align}
which have solution $\ s_1=\frac{16}{149}, s_2=\frac{52}{447},s_3=\frac{46}{149}\ $, so the probability that the last two balls are blue is
$$
\left(\frac{4}{9}\right)\left(\frac{16}{149}\right)+\left(\frac{1}{3}\right)\left(\frac{52}{447}\right)+\left(\frac{2}{9}\right)\left(\frac{46}{149}\right)=\frac{208}{1341}\ .
$$


问:
Why are there repeated patterns in Ulam spiral for multiples?
I had recently, just about a year ago, “discovered” that there are easy to see, clear patterns if you look at the Ulam spiral and just highlight the multiples for numbers.

By Ulam spiral I just mean the layout of spiralling out natural numbers.

These patterns are repeated in x and y direction.

Here are some examples, for multiples of n=22 and n=31

Note that these patterns appear for any scalar and its multipliers. There are no exceptions.

Needless to say, “prime numbers” live in points where some of these patterns appear for the first time, and never again. I have considered that this may be the reason why prime numbers seem to favour “diagonals”, because the multiplier patterns repeat clearly in x/y direction and not chaotically at all - but have not clearly been able to formulate this.

Since the Ulam spiral is based on tracing back the squares of all odd numbers (every bottom-right point - in some orientations this may be the bottom left, - is the next square of an odd number), this means that there are some “formula” F in $(2x+1)^2$ that all resolve to multipliers of a scalar, and these formula are dependent on some constants and $x$ only.

So far, I can only explain this for the pattern that appears at $n = 4$, where I can indeed see a formula in $(2n+1)^2-1$ which resolves to $4n^2+4n$ which obviously generates multipliers of 4.

I have not dared to “do the math” on the other patterns, which are more complex.

Now, a year has gone by and I have been thinking about this a lot, and it is driving me mad.

Could someone help me clear this up so I can let this matter rest?
I wrote a little more about it here - I might have gone into too much assumption in my initial enthusiasm on describing this “discovery” which may ultimately be only a trivial fact in arithmetics.

My question, what could be an explanation for the appearance of these repeated patterns?

https://github.com/buddhabrot/UlamAnalysis/blob/main/README.md

答:
The diagonal, horizontal and vertical lines of the Ulam spiral all have equations of the form $f(n) = 4n^2+bn+c$, as explained on Wikipedia or in this answer. Now say we have
$$
f(n) \equiv r \pmod m
$$
for some $r$ and $m$, then it follows that
$$
\begin{aligned}
f(n+m) &= 4(n+m)^2+b(n+m)+c\\
&= 4n^2+8nm+m^2+bn+bm+c\\
&= \underbrace{4n^2+bn+c}_{f(n)} + \underbrace{m(m+8n+b)}_{\text{multiple of $m$}}
\end{aligned}
$$
which means that
$$f(n+m) \equiv r \pmod m$$
as well. In other words, this is periodic of period $m$, in all directions (horizontally, vertically and diagonally). Finally, the rotation of the pattern in the East, West, North and South quadrants is simply an artefact due to the construction of the spiral itself.

[Edit] A few additional (scattered and maybe not so rigorous) thoughts.

Considering some $m \times m$ block in the Eastern quadrant, for instance
$$\begin{matrix}
\color{red}{30} &\color{red}{55} &88 &129 &178\\
29 &54 &87 &128 &177\\
28 &54 &86 &127 &176\\
27 &52 &\color{red}{85} &126 &\color{red}{175}\\
26 &51 &84 &\color{red}{125} &174
\end{matrix}
$$
it is easy to see that each column must contain exactly one element $\pmod m$ (here $5$). Therefore, leaving aside the regions in the neighborhood of the two main diagonals, every $m \times m$ block will have $m$ marked cells.

Now, in the example block above, I recognized the periodic sequence of quadratic residues $\bmod 5$, which is $\{0,1,4,4,1\}$, which led me to think that the patterns found by the OP are simply (and directly) the sequences of quadratic residues $\bmod m$. But alas, it isn't as simple as that, because we're taking quadratic polynomials modulo $m$. It remains predictable that the patterns will have some form of axial symmetry (due to the evenness of quadratics), and visual hints of parabolic curves shouldn't be too surprising either.

I don't know if much more can be said about the patterns, but I'll keep looking for a more precise characterization.


问:
How to prove a linear map to be the trace function,3x
Let $M_n(F)$ be the matrix space on the field $F$, $f$ be the linear map from $M_n(F)$ to $F$ such that $f(I)=n$, where $I$ is the identitiy matrix. Furthermore, $f(AB)=f(BA)$ for any $A,B\in M_n(F)$. Show that $f=tr$, where $tr(A)=\sum_{i=1}^n a_{ii}$ for $A=(a_{ij}).

I do not have any idea....

答:
For this kind of problems, I would prefer a more elementary approprach. See Nathan Portland's answer, for instance. Here is an answer from a higher perspective.

By the given assumptions, we have $f(AB-BA)\equiv0$. However, the set of all commutators is exactly the space of all traceless matrices (see this question, for instance). So, we have $f(X)=0$ for all traceless matrices $X$. Thus, for a general $X$,
$$
f(X)=\underbrace{f\left(X-\frac{\operatorname{trace}(X)}{n}I\right)}_{=0}+\frac{\operatorname{trace}(X)}{n}\underbrace{f(I)}_{=n} = \operatorname{trace}(X).
$$


问:
Probability of Type 1 Error when using $X_{min}$ as a test static
Let $X_1,X_2, ..., X_{15}$ be a random sample from the exponential distribution with $\lambda > 0 $. To test $H_0 : \lambda = 1/5$ versus $H_A : \lambda < 1/5$ use $X_{min}$ as a test statistic. If $X_{min} \geq 1$ reject the null hypothesis.

My problem is then to compute the probability of a type 1 error. I know that I have to calculate

$$\begin{align*}
  P(\text{Type 1 Error}) & = P(\text{Reject} \ H_0 \ | \ H_0 \ \text{True}) \\
  & = P(X_{min} \geq 1 \ | \ \lambda = 1/5) \\
  & =
\end{align*}
$$

However, I am not sure how to proceed now. As a hint in my book, I have to look an exercise where the PDF for $x_{min}$ is found. Do I have to find the pdf for $X_{min}$ now? I know that the PDF for $x_{min}$
is
$$
f_{min}(x) = n(1 - F(x))^{n-1}f(x)
$$
Do I have to use this? Furthermore, is there any way to calculate this probability with in r studio?

All help is appreciated.

TIA.

答:
You appear to be on track.  Yes, you use that formula for the pdf of $X_{min}$. Books seem to disagree on what to use for the parameter of an exponential.  Does your definition have $\lambda = \mu$ or $\frac{1}{\lambda} = \mu$?  I will use the latter in what follows.

Since $F(x) = 1-e^{-5x}$,

$1-F(x) = e^{-5x}$ and $(1-F(x))^{n-1} = (e^{-5x})^{14} = e^{-70x}$.
So $f_{min} = 14e^{-70x}\cdot 5 e^{-5x} = 70e^{-75x}$
So $P(X_{min} > 1) = \int_{1}^{\infty}70e^{-75x}dx$


问:
Probability of picking a specific card from a deck of X cards after N draws without replacement.
What the title says, what method would I use to calculate the above. To give an example, imagine I wanted to pick the queen of hearts, that specific card, out of a deck of 52 cards. How would I calculate the probability after, say, 20 draws? (without replacement)</p>

答:
All draws are equally likely to show the Queen of Hearts

so the probability of picking it on the $20$th draw is $\frac{1}{52}$

and the probability of picking it in the first $20$th draws without replacement is $\frac{20}{52}$

You can do an induction proof if necessary but I do not think it is.


问:
Existence of L2-functions
On the unit interval (0,1), the functions $f(x)=1/x^p$ are in $L^2$ for $p<1/2$. However, the function $f(x)=x^{-1/2}$ is not in $L^2$ since
$\int_0^{1} \frac{dx}{x}$ = $ln(1) - ln(0)$. And $ln(0)$ is undefined.

But, can someone explain why $f(x)=1/x^p$ is in $L^2$ when p = 1/3.

Here we get $\int_0^{1} \frac{dx}{x^{2/3}} = [\frac{-2}{3}x^{-5/3}]_0^1$ and here we get $\frac {1}{0}$ when we put $0$ and then it goes to infinity. Then why is this function is $L^2$?

Title: Existence of L2-functions

Body: $\int_0^{1}\frac 1 {x^{2/3}}dx= \frac 1 {1-\frac 2  3}x^{1-\frac 2  3}|_0^{1}=3$.


问:
Calculating loan + interest recursively
I am trying to calculate a loan amount where the total principle depends on the payment amount.

Here's a simple example, leaving aside that a) no bank would actually do this, and b) my income is not specified:

I have $0 in the bank right now. I want to take out a loan for a $100 car, for which I will be making a regular annual payment, starting at the end of year 0 with an interest-only payment. My loan rate is 5%. In order to qualify for the loan, the bank wants to make sure that I have a balance equal to twice my annual payment amount throughout the life of the loan.

If I borrow exactly $100, my payment at the end of year 0 would be $5 ($100 x 5%). However, I need to borrow a little extra so that I have a $10 balance in the account (to satisfy the requirement that I maintain a balance of twice my annual payment). This means I actually need a loan of $110. However, a loan of $110 actually raises my payment amount to $5.50, which means I need to maintain a balance of $11 in my account.

This means that I actually need to borrow $111 to begin with, which raises my payment to $5.55, which means I actually need to borrow $111.10, etc.

I have worked it out to a summation that resembles the following, although I can't be certain this is accurate:

$$\sum_{n=1}^x (x-1-n)r^n$$

Where $r$ is the interest rate

I believe this would allow for me to iterate $x$ number of times over a formula that sums the additional principle needed for each incremental amount of interest. However, I'm not sure how to carry this out to infinity.

Ultimately, I need to plug this into Excel as part of a cash flow, but I wanted to think through the math first.

Any ideas?

答:
Assuming that I finally understood the problem correctly, the math is as follows.

Let $M$ be the money you need to spend, $L$ be the money borrowed, $r$ the monthly interest rate, $K$ the ratio of the required balance to the monthly payment.
Your monthly payment (interest only) is $Lr$. You need to have the balance $KLr$ (which you never touch). You have nothing now and have to spend $M$ right away. The leftover from the loan is $L-M$, so we must have $L-M=KLr$, i.e., $L=M/(1-Kr)$.

Of course, this works under the assumptions I described. Other strategies may result in a different answer.

Title: Calculating loan + interest recursively


问:
Clarification required for proof structure in Spivak's Calculus
There is a particular "proof structure" I routinely observe throughout Michael Spivak's Calculus that I hoped I could get some commentary on (I will provide the most recently encountered example shortly).

Such questions will be framed as "Prove $P$", and then Spivak's solutions proceed as follows:

Assume $P$ is true.

Algebraic manipulation

Arrive at a previously proven statement $Q$.

Claim $\square$

Here is one such example:

Prove that $\sqrt{(x_1+y_1)^2+(x_2+y_2)^2} \leq \sqrt{x_1^{\ \ \ 2}+x_2^{\ \ \ 2}} + \sqrt{y_1^{\ \ \ 2}+y_2^{\ \ \ 2}}$

The "$Q$" that we have in our back pocket is the Schwarz inequality, which reads as:

$$x_1y_1+x_2y_2 \leq \sqrt{x_1^{\ \ \ 2}+x_2^{\ \ \ 2}}\sqrt{y_1^{\ \ \ 2}+y_2^{\ \ \ 2}}$$

Here is Spivak's solution (verbatim):

This inequality is equivalent to the squared inequality, $$(x_1+y_1)^2+(x_2+y_2)^2 \leq (x_1^{\ \ \ 2}+x_2^{\ \ \ 2})+ (y_1^{\ \ \ 2}+y_2^{\ \ \ 2})+2\sqrt{x_1^{\ \ \ 2}+x_2^{\ \ \ 2}}\sqrt{y_1^{\ \ \ 2}+y_2^{\ \ \ 2}}$$ which is easily seen to be equivalent to the Schwarz inequality

My major confusion with this proof strategy is that I view it as a proof of the implication $P \rightarrow Q$. If I am keen on proving $P$, certainly I am trying to prove $Q \rightarrow P$, right?

Is Spivak glossing over the fact that in order to work backwards (i.e. to start from the Schwarz inequality and arrive at our initial $P$ statement), we must necessarily be in an algebraic field?

I ask this question because it appears as though one of the unspoken assumptions here by Spivak is that we can casually "reverse" all of the algebraic manipulations that allowed us to traverse from $P$ to $Q$, so that we can then go from $Q$ to $P$.

Is this the correct interpretation?

I believe another relevant assumed-to-be-true statement that must be in play for the reversed $Q \rightarrow P$ direction (at least for this particular example) is that $$x,y \gt 0 \land x^2 \geq y^2 \rightarrow x \geq y$$

答: If Spivak states that he wants to prove $P$, then he observers that $P\iff Q$ and finally he states that $Q$ has been proved before, then since $P\iff Q$, $P$ is proved. There is nothing wrong with that, precisely because we are dealing with an equivalence here.


问:
If $\sup_n \sum_{k=1}^n P(\vert X_{kn} -a_n\vert > b) \leq c$, then can we find a $d$ such that $\sup_n \sum_{k=1}^n P(\vert X_{nk} \vert > d)\leq c$?
This is seperate question in itself but I did start as a follow up question to here: If $\sup_n P(\vert X_n -a_n\vert > b) \leq c$, then can I find a $d \in \mathbb{R}$ such that $\sup_n P(\vert X_n \vert > d) \leq c$?

Let $\{X_{nk} \mid n \in \mathbb{N}, k \leq n \}$ denote a triangular array of real random variables and $\{a_n\}_{n \in \mathbb{N}}$ a sequence of real numbers. Suppose $\max_k P(\vert X_{nk}\vert>\epsilon) \to 0$ for $n \to \infty$ for all $\epsilon>0.$

If $\sup_n \sum_{k=1}^n P(\vert X_{nk} -a_n\vert > b) \leq c$, then can we find a $d \in \mathbb{R}$ such that $\sup_n
 \sum_{k=1}^n  P(\vert X_{nk} \vert > d) \leq c$? And how to choose $d$?

As far as I can tell you need a different solution than the one that Davide provided in the linked question since with $d=\max\{1,d_1,\dots,d_{n_0 -1}\}$ would yield the bound
$$
\sum_{k=1}^n  P(\vert X_{nk} \vert > d) \leq nc
$$
which doesn't give anything useful when applying $\sup_n$ on both sides.

答:
Note that

$$\sum_{k=1}^n P(|X_{nk}-a_n|>b)\geq \sum_{k=1}^n P(|X_{nk}|>b+|a_n|)$$

So you get the result immediately if $\sup_n |a_n|<\infty$. However, we must have $\limsup_{n\to\infty} |a_n|\leq b$.

Indeed, let $|A|>b$. Then, pick $N$ so large that $\max_k P(|X_{nk}|>|A|-b)\leq \frac{1}{2}$ for $n\geq N$. Then, for such, $n$, we have

$$
\sum_{k=1}^n P(|X_{nk}-A|>b)\geq \sum_{k=N}^n P(|X_{nk}-A|>b)\geq \sum_{k=N}^n P(|X_{nk}|< |A|-b)\geq \frac{(n-N)}{2},
$$
from which it immediately follows that $\limsup_{n\to\infty} |a_n|< b$. But this implies that $\sup_n |a_n|<\infty$ and thus, $d:=b+\sup_n |a_n|$ works.


问:
Choosing $a$ s.t. $\frac{a^k - 1}{a-1}$ is not a prime power

Let us suppose that we are presented with a positive integer $k$ and asked to come up with a positive integer $a$ such that $\frac{a^k - 1}{a-1}$ is not a prime power, or just to prove in an elementary way that this can be done. How do we proceed?

Of course, in practice this is essentially trivial. But is there a cannonical choice? For that matter, is there an elementary proof that shows that this is always possible?

This came up in a discussion with another Brown mathie about interesting things to talk about in an elementary number theory class I will be teaching this summer. To be more specific, we were trying to prove the infinitude of primes in arithmetic progressions of the form $1, 1+p^n, 1+2p^n, ...$ using a naive a completely elementary approach (hopefully to be replicated by my students this summer), and this side topic came up.

答:
Let $a=2k+1$ then
$$
\frac{a^k-1}{a-1} = \frac{\sum_{i=1}^k\binom{k}{i}(2k)^i}{2k}\equiv k\pmod{k^2}
$$
which is not $k$, divisible by $k$ but not $k^2$, and hence not a prime or a power for $k>1$.</p>


问:
Why is the definition of the element $rv$ of the tangent space chart-independent?
Let's assume we have a $m$-dimensional smooth manifold $\mathscr{M}$. We can define the tangent space $T_p\mathscr{M}$ at a point $p \in \mathscr{M}$ as the set of equivalence classes of curves in $\mathscr{M}$ going through $p$. Then to show that $T_p\mathscr{M}$ is a vector space my textbook defines $r v$ as

$$ r v := [\phi^{-1} \circ (r\phi \circ \sigma)] \, ,$$
where $r \in \mathbb{R}$, $v = [\sigma]\in T_p\mathscr{M}$ and $\phi$ a chart around $p$ satisfying $\phi (p) = 0$.

I was able to prove this definition doesn't depend on the choice of representative $\sigma$, but how do you proceed to prove that it doesn't depend on the choice of chart $\phi$?

答:
I will try to outline the main ideas.

First, note that $\phi(U)$ being an open subset of $\mathscr{M}$ it is itself a smooth manifold of dimension $m$. So, instead of looking at $T_p\mathscr{M}$ one can look at $T_0(\phi(U))$ and the main advantage there is that a chart is not required anymore to define the vector space structure on $T_0(\phi(U))$. Indeed, $\phi(U)$ already being a subset of $\mathbb{R}^m$ a chart is not needed to push the values taken by the curves into $\mathbb{R}^m$ to use the vector space structure thereof.

The vector space structure on $T_0(\phi(U))$ is defined as follows.

$v_1 + v_2 = [\sigma_1] + [\sigma_2] = [\sigma_1 + \sigma_2]$

$rv = r[\sigma] = [r\sigma]$

Then, one can check that the vector space $T_0(\phi(U))$ is isomorphic to $\mathbb{R}^m$ with the isomorphism mapping $v \in \mathbb{R}^m$ to $[\sigma_v]$, where $\sigma_v(t) := t v$.

Moreover, we have a bijection $\phi_*$ from $T_p\mathscr{M}$ to $T_0(\phi(U))$ mapping $v := [\sigma]$ to $[\phi \circ \sigma]$. One can use this bijection to transport the vector space structure on $T_0(\phi(U))$ to $T_p\mathscr{M}$ making $\phi_*$ into an isomorphism of vector spaces. The resulting vector space structure on $T_p\mathscr{M}$ is precisely the one defined partially in your post.

Finally, given another chart $(V, \psi)$ with $p \in V$ and $\psi(p) = 0$, one can define in the same way a vector space structure on $T_p\mathscr{M}$ using $\psi_*$.
But what has been shown above is that both vector space structures are the same, i.e. they are isomorphic, both being isomorphic to $\mathbb{R}^m$ (through $T_0(\phi(U))$ and $T_0(\psi(V))$, respectively).


问:
Phone numbers expressed as operations of numbers with less digits
Where I live, mobile phone numbers tend to be combinations of 9 digits, mainly starting by 6, and I've been days wonder whether is it possible to express a given number as sums/products involving less digits than the original, and if is it possible to conceive a reasonable algorithm to find a solution.

My first thought was to factorize it, but prime factors usually involve some relatively big prime number.

I've managed to find nice examples testing some combinations. For instance, let's say my phone number is 671088640. This number can be obtained as 5*8^9, that is, you only need to remember 3 digits (and 2 operators).

How would you approach this problem away from brute force?

Thank you.

答:
I will assume the operations you allow are only +/-, × and ^. My reasoning for that is, that there is noting inherently special about the operations that "have a name" and if I would allow ANY operation, then for the phone number T I could just take the operation
a#b = abT or even a#b = T for any a,b.

From those 3 operations only ^ saves on digits: + increases the number of digits required and * leaves it the same. This means, that our goal will be to find a way to incorporate a^b for b as big as possible into our equation.

As pointed out before, it will rarely be the case, that T is divisible by a high power of a number. However, when combining this with +/- there is hope for one of the numbers
T, T+1, T-1, T+2, T-2 etc to be divisible by a high power of a number. Unfortunately, this will still not help for most numbers.

When you look at the information theoretical side of it, you will see, why there can be no fool-proof algorith for this:

Using 9 decimal points, I can represent 10^9 different numbers.
Using sequences of 0,1,2,3,4,5,6,7,8,9,+,-,^,× of lenght at most 7 we can represent less than (as some sequnces represent the same number) 14^7 < 10^9 different numbers. This means, that you cannot hope for a general algorithm to significantly (or probably at all) cut down on the number of digits you have to remember. (And now you also have to remember signs)

For a specific number T your best bet is to hope that one of
T-99, ..., T+99 is a multiple of a high prime power, or that you can find small a, b, k, l such that
T = a×8^k + b×9^k
or, if that fails try 7^k, ...


问:
Map of spectra induced by $\mathbb{C}[t] \to \mathbb{C}[t]$, $t \mapsto t^2$ (Vakil's Rising Sea, 3.2.O)
In Vakil's ‘Rising Sea’, exercise 3.2.O goes (edited for brevity):

Consider the map of complex manifolds sending $\mathbb{C} \to \mathbb{C}$ via $x \mapsto y = x^2$. Interpret the corresponding map of rings as given by $\mathbb{C}[y] \to  \mathbb{C}[x]$ by $y \mapsto  x^2$.

Here is my solution:

Consider $f: \mathbb{C}[y] \to  \mathbb{C}[x]$ where $f(y) = x^2$. Then the induced map $f^{\ast}: \mathrm{Spec} \, \mathbb{C}[x] \to  \mathrm{Spec}\, \mathbb{C}[y]$ is given by $[\mathfrak p] \longmapsto [f^{-1}(\mathfrak p)]$. We have to prove that $f^\ast((x-a)) = (y-a^2)$. Since $f(y-a^2) = (x-a)(x+a) \subseteq (x - a)$, we have one side of the inclusion.

Now suppose $q(y) = \sum b_k y^k \in \mathbb{C}[y]$ and $f(q(y)) = \sum b_k x^{2k} = (x-a)p(x)$ for some polynomial $p(x) \in \mathbb{C}[x]$. Since $\sum b_k a^{2k} = (a-a)p(a) = 0$, we have $\sum b_k (a^2)^k = 0$, and thus $q(a^2) = 0$ and $q(y) \in (y-a^2)$.

Is my proof correct? And how could I improve it?

答:
I think your proof works.

Alternatively you can argue as follows: The ideal $(y-a^2) \subset \Bbb C[y]$ is a maximal ideal, so $y-a^2 \in f^{-1}(\mathfrak p)$ shows $(y-a^2) = f^{-1}(\mathfrak p)$, since $f^{-1}(\mathfrak p) \neq \Bbb C[y]$.


问:
An endomorphism of a field is the identity on a subfield. Is it an automorphism?
Let $K$ be a subfield of a field $L$, both infinite. Let $\phi$ be an endomorphism on $L$ such that $\phi|_{K} = id|_{K}$

Is it true that $\phi$ is an automorphism?

I believe, it is so if $L$ is a finite extension of $K$.
A counter example for finite $K$ and infinite $L$ would be the Frobenius's endomorphism on $\mathbb{Z}_2(t)$, which is identity on $\mathbb{Z}_2$, but not surjective.

答:
Here is a variant of the argument given by @egreg.

A ring homomorphism between fields that preserves the multiplicative identity is necessarily injective because there are no nontrivial ideals in a field.

Therefore, $\phi$ is an injective $K$-linear transformation and so is surjective when $L$ is a finite extension of $K$.


问:
If $f : X \to Y$ is a homeomorphism and $\rho(x,y)= d'(f(x),f(y))$ defines a metric in $X$. Show that $d$ and $\rho$ are equivalent.

Let $(X,d)$ and $(Y,d')$ be metric spaces and $f : X \to Y$ a homeomorphism. Show that $\rho(x,y)= d'(f(x),f(y))$ defines a metric $\rho$ in $X$ and that $d$ and $\rho$ are equivalent.

To show that $\rho(x,y)$ is a metric I have $$\rho(x,y)=d'(f(x),f(y)) \leqslant d'(f(x),f(y)) + d'(f(y),f(z)) = \rho(x,y)+\rho(y,z) \\ \rho(x,y)=d'(f(x),f(y))=d'(f(y),f(x)) = \rho(y,x) \\  \rho(x,y)=0 \iff d'(f(x),f(y))=0 \implies f(x)=f(y) \underset{f \text{ homeo}}\iff x=y$$

thus $\rho$ is a metric in $X$. To show that $\rho$ and $d$ are equivalent I'm  a bit lost. I have three metrics here and I'm not sure where I should even use $d$ since $\rho$ depends on $d'$ not $d$.

In essence I would need to show that $\operatorname{id} :(X,d) \to (Y,d') \to  (X,\rho)$ is
homeomorphic? The fact that $f$ is homeomorphic implies that there exists and continuous inverse $f^{-1}:(Y,d') \to (X,d)$, but this also says nothing about $(X,\rho)$.

$\textbf{If}$ I would have that $f^{-1}:(Y,d') \to (X,\rho)$, then picking $y_1,y_2 \in Y$ I would have that $$\rho(f^{-1}(y_1), f^{-1}(y_2))=d'(f(f^{-1}(y_1)),f(f^{-1}(y_2)) = d'(y_1,y_2)$$ which would imply that $f^{-1}$ is a bijective isometry from $(Y,d') \to(X,\rho)$ and thus homeomorphism. Furhtermore I would have that $f^{-1} \circ f :(X,d) \to(X,\rho)$ would be a homeo since it's a composition of two homeos. So $\operatorname{id} :(X,d) \to  (X,\rho)$  is homeomorphic and $d$ and $\rho$ are equivalent.

I'm very confused about the fact that I have three metrics. Any clarification would be welcome.

答:
Proof 1

For any $x \in X$, $D$ a distance on $X$ and $r \gt 0$, denote by $B_r(x;D)$ the open ball for the distance $D$ centered on $x$ and of radius $r$.

By definition of equivalent distances, we have to prove that for any $x \in X$ and $r \gt 0$, we can find $r_1,r_2 \gt 0$ such that

$$B_{r_1}(x,d) \subseteq B_{r}(x,\rho) \text{ and } B_{r_2}(x,\rho) \subseteq B_{r}(x,d).$$

By definition of the distance $\rho$, we have

$$B_{r}(x,\rho)= \{y \in X \mid d^\prime(f(x),f(y)) \lt r\} .$$ As $f$ is supposed to be an homeomorphism between $(X,d)$ and $(Y,d^\prime)$, it is continuous at $x$ and it exists $r_1 \gt 0$ such that for all $y \in X$ with $d(x,y) \lt r_1$, we have $d^\prime(f(x),f(y)) \lt r$. This exactly means that $B_{r_1}(x,d) \subseteq B_{r}(x,\rho)$.

I let you deal with the case of finding $r_2$ using the continuity of $f^{-1}$.

Proof 2

Consider

$$\operatorname{id} :(X,d) \mathop{\to}^{f} (Y,d^\prime) \mathop{\to}^{f^{-1}}  (X,\rho).$$

$\operatorname{id}$ is well defined as $f$ is supposed to be a homeomorphism between $(X,d)$ and $(Y,d^\prime)$.

$f^{-1}$ is an isometric map between $(Y,d^\prime)$ and $(X, \rho)$ as for $x^\prime = f(x), y^\prime =f(y) \in Y$ we have
$$\rho(f^{-1}(x^\prime),f^{-1}(y^\prime))=\rho(x,y)=d^\prime(f(x),f(y))= d^\prime(x^\prime,y^\prime)$$

$\operatorname{id}$ which is the composition of two continuous maps (the homeomorphism $f$ and the isometry $f^{-1}$) is $(d,\rho)$-continuous.

In a similar way, we could prove that

$$\operatorname{id} :(X,\rho) \mathop{\to}^{f} (Y,d^\prime) \mathop{\to}^{f^{-1}}  (X,d)$$ is $(\rho,d)$-continuous.

Finally $d, \rho$ are equivalent as $\operatorname{id}$ is both $(d, \rho)$ and $(\rho,d)$ continuous.


问:
At $z=0$ the function $f(z)=\exp({z\over 1-\cos z})$ has
At $z=0$ the  function $f(z)=\exp({z\over 1-\cos z})$ has

$1$. A removable singualrity

$2$. A pole

$3$. An essential singularity

$4$. Laurent series around $z=0$ has infinitely man positive and negative power of $z$

I see $\lim_{z\to\infty}f(z)=\infty\cup -\infty$ if you approach to $0$ from left and right side of the real line, so $f$ can not be bounded near $0$ so it has $f$ essential singularity?

答: How do you know it isn't a pole? For small $z$, $\frac{z}{1-\cos z}\approx \frac{z}{z^2}=\frac{1}{z}$, so that for small $z$ your function is $\displaystyle  e^{\frac{1}{z}}=\Sigma \frac{1}{n!z^n}$, and this clearly goes to infinity faster than any polynomial goes to $0$, ie. there is no $m$ such that $\,\,\displaystyle  z^m\Sigma \frac{1}{n!z^n}$ is bounded near $0$, hence it is an essential singularity.</p>


问:
Which group is C0?
This is likely a silly question, but I cannot find an answer on the internet. I used Sage-math to find the structure of a group presented as:

$$\langle x,y; xy=1 \rangle$$

It returned C0: which group is it?

EDIT Here is the code I used

```<code>sage: G.<x,y,z,t>=FreeGroup()
sage: H=G.quotient([x*y,x*z,x*t])
sage: H.structure_description()
'C0'
</code>```

答:
I'm not sure what $\;C_0\;$ means, but that group of yours is $\;\Bbb Z\;$ , the infinite cyclic group (though you wrote it multiplicatively). Observe that from the unique relation given, we get

$$xy=1\implies y=x^{-1}\implies K:=\langle x, y\;|\; xy=1\rangle=\langle x,x^{-1}\rangle =\langle x\rangle\cong\Bbb Z$$

Title: Which group is C0?


问:
Calculating force on the side wall of a container
My daughter is busy with Calculus AB and she asked me about a problem that she was having. In one of her multiple choice quizzes, the examiner asked about the force on the side of a half full cylinder that is lying on it's side. The cylinder is 20 feet long and has a diameter of 12 feet. The fluid has a density of 50.

The cylinder would look something like this:

My daughter got the right answer but only when she ignored the length of the cylinder. That seems wrong to me. If a dam is 1 foot wide then the force of the water on the dam wall would be a lot less than it would be for a dam that was 10000 feet wide.

It's been a while but looking up the formula in her notes says that it should be this:

$$
\int_a^b \rho \cdot h(y) \cdot L(y) dy
$$

The formula for the circle is:

$$
x^2+(y-6)^2=6^2
$$

In terms of x that would be:

$$
x=\sqrt{6^{2}-\left(y-6\right)^{2}}
$$

h(y) is the depth of the slice. The L(y) dy is the area of each slice. In this case, the slice is 20 long and has a width of 2 times x (the formula above).

which is:

$$
\int_0^6 \biggl( 50 \cdot (6-y) \cdot 20 \cdot 2 \cdot \sqrt{6^{2}-\left(y-6\right)^{2}}\biggl) dy
$$

Is that correct or is the length of the cylinder supposed to be ignored? Is the area a 2d slice that only includes the side wall itself or is the area a slice that includes a full horizontal slice of the container?

答: The area under pressure in question does not depend on the length of the cylinder. Think about a diver, the pressure only depends on the depth of submersion. It does not matter whether the diver is in a tank of water or in the ocean. The length would be important if we needed to find force acting on the side of the tank.</p>


问:
Maximum of norm inducted by inner product
I want to choose such two norms $\Vert\cdot\Vert_a, \; \Vert\cdot\Vert_b$ for which:

$$\Vert\cdot\Vert_c := \max\{\Vert\cdot\Vert_a, \; \Vert\cdot\Vert_b\}$$

Is a norm inducted by inner product.

My work so far

According to Maximum of two norms is norm I'm asured that my object is well defined. Norm is inducted by inner product when the following formula is satisfied:

$$\Vert x+y \Vert^2 + \Vert x-y \Vert^2 = 2(\Vert x \Vert^2 + \Vert y \Vert^2)$$

Let's define $A := (x_1, y_1)$, $B:= (x_2, y_2)$ and rewrite the formula above:

$$\Vert A+B\Vert_c^2 + \Vert A-B \Vert_c^2 = 2(\Vert A\Vert_c^2 + \Vert B \Vert_c^2)$$

$$\Vert (x_1+x_2, y_1+y_2)\Vert_c^2 + \Vert (x_1-x_2, y_1-y_2)\Vert_c^2 = 2(\Vert (x_1, y_1)\Vert_c^2+\Vert (x_2, y_2)\Vert_c^2)$$

$$ \max{\{\Vert x_1 + x_2 \Vert_a, \Vert y_1 + y_2 \Vert_b\}}^2 + \max{\{\Vert x_1 - x_2 \Vert_a, \Vert y_1 - y_2 \Vert_b\}}^2 = 2( \max{\{\Vert x_1 \Vert_a, \Vert y_1 \Vert_b\}}^2 + \max{\{\Vert x_2 \Vert_a, \Vert y_2 \Vert_b\}}^2)$$

And here I wasnt sure which norms $a, b$ to chose. I tried some combinations $\mathbb{l}_1$ with supremum but it doesn't give me more than just bizzare looking expressions.

Could you please help me findin some norms $a$ and $b$ for which parallelogram law holds ?

答:
Here is a different idea. The unit ball of the $\|\cdot\|_c$ norm is equal to the
intersection of the unit balls of the $\|\cdot\|_a$ and $\|\cdot\|_b$ norms. Now choose two norms such that the intersection of their unit balls is the Euclidean unit ball.

Here is one on $\mathbb R^2$:
$$
\|(x_1,x_2)\|_a:= \begin{cases} \sqrt{x_1^2 + x_2^2} & \text{if }x_1x_2 \ge 0\\\max(|x_1|,|x_2|) & \text{if }x_1x_2 < 0.
\end{cases}
$$
Define $\|(x_1,x_2)\|_b:=\|(-x_1,x_2)\|_a$. Since $\max(|x_1|,|x_2|)\le\sqrt{x_1^2 + x_2^2}$, we have $\max(\|(x_1,x_2)\|_a,\|(x_1,x_2)\|_b) = \sqrt{x_1^2+x_2^2}$.


问:
Derivative of $f(x) = \sqrt[3]{(x-1)(x-2021)}|\sin(\pi x)|$ at $x = 1$
So I got this really strange contradiction that I couldn't explain to myself and I'd really love some clarification on the pitfalls that I'm potentially facing.

Let $$f(x) = \sqrt[3]{(x-1)(x-2021)}|\sin(\pi x)|$$

The very strange thing that's happening is that when I try to evaluate $f'(1)$ with the direct definition as:
$$f'(1) = \lim_{x\to 1}\frac{f(x) - f(1)}{x - 1} = \lim_{x\to 1}(x-1)^{-\frac{2}{3}}(x-2021)^\frac{1}{3}|\sin(\pi x)| = 0$$

But when I try to explicitly find $f'(x)$ as an expression for $x$:
$$f'(x) = \frac{(4x-4044)|\sin(\pi x)|^2 + 3\pi(x-2021)(x-1)\sin(2\pi x)}{6\left((x-1)(x-2021)\right)^\frac{2}{3}|\sin(\pi x)|}$$
and then plug in $x = 1$, I clearly get a $0$ in the denominator, hence $f'(1)$ is undefined.

There's obviously something smelly, but I find this one to be a really interesting case that I couldn't explain to myself. Any advice is welcome and appreciated! Thanks in advance!

答:
$0$ is the correct derivative, which you got in the first calculation. I assume however that you overlooked something in the calculation, as the limit is far from simple.

When evaluating the derivation limit directly, you correctly get that it is the limit of a product of 3 terms:

$$f'(1)=\lim_{x\to 1}(x-1)^{-2/3}(x-2021)^{1/3}|\sin(\pi x)|$$

Now the 'easy' solution to this would be if we could calculate the limit of each term directly, then multiply their results. But it turns out this doesn't work, as one term tends to infinity:

$$\lim_{x\to 1}(x-1)^{-2/3} = \infty,$$
$$\lim_{x\to 1}(x-2021)^{1/3} = \sqrt[3]{-2020},$$
$$\lim_{x\to 1}|\sin(\pi x)| = 0,$$

the last 2 limits calculated simply be realizing that the functions taken to the limit are continuous at the given point. So one limit is $\infty$, one is $0$ and one is a finite, non-zero number. That means the limit of their product could be anything: $0$, some other number, $\infty$ or "diverging" in the strict sense (not tending to a finite number or infinity). We need to evaluate the limit of the product

$$\lim_{x\to 1}(x-1)^{-2/3}|\sin(\pi x)|,$$

and whatever we find as a result (if there is one at all), it will just have to be multplied by $\sqrt[3]{-2020}$ in order to get the derivative.

Now we have

$$(x-1)^{-2/3}|\sin(\pi x)| = \frac{|\sin(\pi x)|}{(x-1)^{2/3}} =  \frac{|\sin(\pi x - \pi)|}{(x-1)^{2/3}} = \frac{|\sin(\pi(x - 1))|}{(x-1)^{2/3}} = (x-1)^{1/3}\frac{|\sin(\pi(x - 1))|}{(x-1)} = \pi(x-1)^{1/3}\frac{|\sin(\pi(x - 1))|}{\pi(x-1)}.$$

These are mostly simple equivalent transformations, where the fact that $|\sin(x)|$ has period $\pi$ has ben used for the second equality sign.

Now if you've seen a bit of trigonometric limits, the last term contains a part that is very near to what you may have seen before, if you set $y=\pi(x-1)$

$$\lim_{y\to 0}\frac{\sin(y)}y = 1$$

The only difference is that our term contains an absolute value. But that absolute value doesn't change that

$$\frac{|\sin(\pi(x - 1))|}{\pi(x-1)}$$

is bounded around $x=1$. So we see that

$$\pi(x-1)^{1/3}\frac{|\sin(\pi(x - 1))|}{\pi(x-1)}$$

is the product of 2 factors, the first of which ($\pi(x-1)^{1/3}$) tends to $0$ when $x\to 1$ and the second ($\frac{|\sin(\pi(x - 1))|}{\pi(x-1)}$) stays bounded when $x \to 1$. That means the product of those 2 functions tends to $0$ as well:

$$\lim_{x \to 0}\pi(x-1)^{1/3}\frac{|\sin(\pi(x - 1))|}{\pi(x-1)} = 0.$$

That finally means that

$$f'(1)=\lim_{x\to 1}(x-1)^{-2/3}(x-2021)^{1/3}|\sin(\pi x)| = \lim_{x \to 0}\pi(x-1)^{1/3}\frac{|\sin(\pi(x - 1))|}{\pi(x-1)} \times \lim_{x\to 1}(x-2021)^{1/3} = 0 \times \sqrt[3]{-2020}=0.$$

Now what happened when using the second method to calculate the derivative explicitly?

The function $f$ is given as the product of 2 functions, and it turns out that neither factor is differentiable at $x=1$. You can do the calculations yourself (you did already), but the Wolfram alpha picutures show that the root part has an "infinite" derivative:

The $\sin$ part OTOH has different left and right derivates (due to the absolute value function "flipping" the sin function at $x=1$):

So you simply can't use the product rule for the function at the point $x=1$, because that requires that both functions have a derivative at that point, which isn't true. One would guess that the product of 2 such functions would then also have no derivative for $x=1$, but the above calculation shows that is not true.

Finally, here comes the picture of how the function $f$ looks like around $x=1$:

It looks continuous, though one can't really see what the derivative at $x=1$ is, but that cleared up by the above calculation.