问:
Nonlinear first-order differential equation with a simple parametric solution.
I have to solve the nonlinear first-order differential equation
$$\frac{a-y'}{\sqrt{1+y'^2}}e^{-a \arctan y'}=bx+c,$$
where $a,b,c$ are constants, and $y$ is a function of $x$.

Obviously, there is no way to put it in the form $y'=f(x)$ and integrate. However, I know that a simple parametric solution exists:
$$\begin{align}
x(t) &= c_1 + c_2 e^{-a t}(a\cos t-\sin t) \\
y(t) &= c_3 + c_2 e^{-a t}(a\sin t+\cos t),
\end{align}
$$
for some constants $c_1,c_2,c_3$ (that depends on $a,b,c$). How can we arrive at this  solution? What is the method?

答:
Here is a start, let

$$ \arctan(y')=t \implies y' = \tan(t). $$

Subs back in the ode, we have

$$ \frac{a-y'}{\sqrt{1+y'^2}}e^{-a \arctan y'}=bx+c \implies \frac{a-\tan(t)}{\sec(t)}e^{-a t}=bx+c $$

$$ \implies  x(t) = -\frac{c}{b} + \frac{1}{b}(a\cos(t)-\sin(t))e^{-at}  $$

$$ \implies  x(t) = c_1 + c_2e^{-at}(a\cos(t)-\sin(t)).  $$

Now, you need to find $y(t)$. I think you can finish the task.


问:
In a locally compact Hausdorff space, a subset $A$ is closed if it has closed intersection with every compact subset of $E$
Let $(E,\tau)$ be a locally compact Hausdorff topological space, and $A \subseteq E$ some set on $E$, such that $A$ has closed intersection with each compact subset of $E$. I am asked to show that $A$ is closed.

The definition for locally compact I am provided is the following: A topological space is called locally compact provided each point in it has a neighborhood contained in some compact set.

Any help would be very appreciated.

答:
Let $A\subseteq E$ have the property that $A\cap K$ is closed in $K$ for all compact $K\subseteq E$. We want to show that $A$ is closed whenever $E$ is locally compact Hausdorff, so we will show that $E\setminus A$ is open.

Let $x\in E\setminus A$, let $K$ be a compact neighbourhood of $x$, and let $U\subseteq K$ be an open neighbourhood of $x$. By hypothesis $A\cap K$ is closed in $K$, so $K\setminus A$ is open in $K$ which means that $K\setminus A\cap U$ is open in $U$ and thus in $E$, so $K\setminus A\cap U$ is a neighbourhood of $x$ disjoint from $A$, showing that $A$ is closed.

As an aside spaces in which a set is closed iff its intersection with all compact sets is closed are called $k$-spaces or compactly generated spaces. The above shows that locally compact Hausdorff spaces are compactly generated, and it turns out that being compactly generated is equivalent to being the quotient of a locally compact Hausdorff space.

ContentLicense: CC BY-SA 4.0


问:
How to prove $\mathbb{Z}[i]$ is a ring but not a field?
Let i be the imaginary unit of $\Bbb{C}$ and $\mathbb{Z}[i]:= \{a + ib, ab \in\mathbb{Z} \} $

So the task is to prove that this a ring regarding multiplication and addition in $\mathbb{C}$ which is commutative, free of zero divisors and posseses a 1 element.
I already found out that it is a ring and that the 1 Element is (1,0) and 0 Element is (0,0), so in both cases $i= 0$. So I guess this is like proving a trivial Ring.
I am just confused because I am not sure about the $i$. How would I complete the proof? I am stuck because I don't know how to prove that it is free of zero divisors and not a field.
Thank you for the help.

答: The map $\Bbb Z[i]\to \Bbb Z/2\Bbb Z$, $a+bi\mapsto a+b+2\Bbb Z$ is a non-trivial ring homomorphism with non-trivial kernel. Such a thing does not exist for fields.


答:
Consider the norm $|a + bi| = \sqrt{a^2 + b^2} \geq 1$ for $a + bi \in \mathbb{Z} \setminus \{0\}$. This is multiplicative (it is complex norm).

Hence, the element $1 + 2i$ with $|1 + 2i| = \sqrt{5}$ can't be invertible. If it had an inverse $x$, then $|x| \sqrt{5} = | 1 | = 1$, but norms in the non-zero Gaussian integers are at least $1$.

Title: How to prove $\mathbb{Z}[i]$ is a ring but not a field?

Body:
Let i be the imaginary unit of $\Bbb{C}$ and $\mathbb{Z}[i]:= \{a + ib, ab \in\mathbb{Z} \} $

So the task is to prove that this a ring regarding multiplication and addition in $\mathbb{C}$ which is commutative, free of zero divisors and posseses a 1 element.
I already found out that it is a ring and that the 1 Element is (1,0) and 0 Element is (0,0), so in both cases $i= 0$. So I guess this is like proving a trivial Ring.
I am just confused because I am not sure about the $i$. How would I complete the proof? I am stuck because I don't know how to prove that it is free of zero divisors and not a field.
Thank you for the help.


问:
Is there a model exists, such that homomorphic to the set of real numbers but the order is given by $\in$?
The question is motivated by the construction of a natural number system, where its order is given by $\in$, so is there any analogous exists for the set of real numbers?

It seems impossible for me since the cardinality of the set of real numbers is $2^{\aleph_0}$, but I am not really sure about this. To make a question precise, is there any model (with some order) $\mathcal{M}$ such that $\mathcal{M}\models T$ where $T$ is the theory of real numbers, and the order on $\mathcal{M}$ is given by $\in$?

Sorry if I am writing with ignorance with mathematical logic, I know very little about the topic.

答: The membership relation is well founded, which lends itself particularly well to being an ordering relation on the naturals; but the ordering relation on the reals is definitely not well founded. So your hunch is correct, this can't be done.</p>


问:
entire bijection of $\mathbb{C}$ with 2 fixed points
Besides the identity map, is there an entire function $f$ that is a bijection from $\mathbb{C}$ to $\mathbb{C}$ and has 2 fixed points? Thank you for the help.</p>

答: No. Let $f$ be an entire function, and consider the singularity of $f$ at infinity. If it's removable, then by Liouville's theorem $f$ is constant and so not a bijection. If it's essential, then by Picard's great theorem $f$ is not injective. If it's a pole, then $f$ is a polynomial. If the degree of $f$ is greater than one, then $f$ is not injective. So $f$ is a linear polynomial, and the only linear polynomial that fixes two points is the identity.</p>

ContentLicense: CC BY-SA 3.0


问:
What is the chance that an infinite series of random numbers will, at some point, repeat all previously generated numbers in that order?
The main reason of this post is to provide a writeup of what I think is a correct solution to this problem from the r/math subreddit. I'll quote the poster to phrase the problem:

What is the chance that an infinite series of random numbers will, at some point, repeat all previously generated numbers in that order?

While there was some confusion in the comments, I believe what the poster means is something like the sequences below:
$$7,7,...$$
$$3,2,3,2,...$$
$$7,8,6,7,8,6,...$$
Where the dots afterward are arbitrary as the condition has already been fulfilled. If we start writing out a sequence of digits in this way, where the probability of selecting any digit is equal, what is the probability that something like this will happen?

I'll post my solution below. Feel free to comment or correct me.

答:
EDIT: This answer is incorrect, but I'll leave it up as it was my original approach.

First, some definitions. Let the digits of the sequence be defined as the list $a_0,a_1,a_2,...$ Let the blocks of $2^n$ digits be defined as $b_1,b_2,...$ What I mean by this is more easily illustrated by looking at the decimal expansion of $e$:
$$ \underbrace{2.7}_{b_0}\underbrace{18}_{b_1}\underbrace{2818}_{b_2}\underbrace{28459045}_{b_3}\underbrace{2353602874713526}_{b_4}...$$
$b_n$ is the $n$th block of $2^n$ numbers, save for the first one. I will also define $c_n$ as the $n$th partial sequence - eg for $\pi$,
$$c_5=3,1,4,1,5$$
With that out of the way, the solution is rather simple. Let the probability of such a sequence occurring be $p$. Then we can write
$$p=\Pr(a_1=a_0)+\Pr(b_1=c_2)\Pr(a_1\neq a_0)+\Pr(b_2=c_4)\Pr(b_1\neq c_2)+\Pr(b_3=c_8)\Pr(b_2\neq c_4)+...$$
$$p=0.1+0.01\cdot0.9+\sum_{n=2}^\infty \Pr(b_n=c_{2^n})\Pr(b_{n-1}\neq c_{2^{n-1}})$$
$$p=0.109+\sum_{n=2}^\infty \left(\frac{1}{10}\right)^{2^n}\left(\frac{9}{10}\right)^{2^{n-1}}$$
$$p\approx 0.1090810065610000431...$$

Perhaps someone can find a closed form for the above sum.


答:
There are no non-zero zero-divisors because $\mathbb{Z}[i]$ is a subring of $\mathbb{C}$, which is a field. Or, if you prefer:
$$ (a+bi)(x+yi)=(ax-by)+(bx+ay)i.$$
So the product is equal to zero if and only if $ax-by=bx+ay=0$. Try to deduce that either $a=b=0$ or $x=y=0$.

If $\mathbb{Z}[i]$ were a field, then every non-zero element would have a multiplicative inverse. Let $\alpha$ be an inverse of $2$. But $\mathbb{Z}[i]$ is a subring of $\mathbb{C}$, so $\alpha \in \mathbb{C}$, and inverses are unique, so $\alpha = 1/2$. However, $1/2 \notin \mathbb{Z}[i]$, contradiction.


问:
functions and infinite intersections
let $f$ be a function.

If it is given that $$ f\left[\bigcap_{a\in A}F_{a}\right]\subseteq\bigcap_{a\in A}f[F_{a}]$$

then, if it is further given that $f$ is one to one, prove that the $\subseteq$ can be replaced with an =.

I am fairly certain that my proof is wrong, as I am useless at proofs. :(
Here is what I have :

let $$x \in \bigcap_{a \in A}f[F_{a}]$$

then
$$ x \in f[F_{a_{1}}] \text{  and} x\in f[F_{a_{2}}] ...$$
$$ \text {  so  }f(x) \in F_{a_{1}} \text{  and} x \in F_{a_{2}} $$
$$ f(x) \in \bigcap_{a \in A} F_{a} $$
$$ x \in f\left[\bigcap_{a \in A} F_{a}\right] $$

please help me figure this out.

答:
First a minor technical objection: the index set $A$ need not be countable, so you can’t assume that you can list it as $a_1,a_2,\dots\,$.

Your major error comes when you go from $x\in f[F_a]$ to $f(x)\in F_a$: $x\in f[F_a]$ means that there is some $y\in F_a$ such that $x=f(y)$. However, since you’re assuming that $f$ is one-to-one, this $y$ is unique and is simply $f^{-1}(x)$. Thus, from $x\in F_a$ you can infer that $f^{-1}(x)\in F_a$. You can do this for each $a\in A$, so you can now conclude that $f^{-1}(x)$ belongs to every one of the sets $F_a$ with $a\in A$ and hence that

$$f^{-1}(x)\in\bigcap_{a\in A}F_a\;.$$

Now apply $f$ to both sides to conclude that

$$x=f\left(f^{-1}(x)\right)\in f\left[\bigcap_{a\in A}F_a\right]\;,$$

which is exactly what you wanted.

Title: functions and infinite intersections


答:
The axiom of foundation tells us that $\in$ is a well-founded relation, i.e., there are no infinite descending $\in$-chains in the universe of sets. Since $\mathbb{R}$ (and any model of the theory of $\mathbb{R}$) contains infinite descending $<$-chains, this gives a negative answer to your question. More precisely: if a set $M$ is linearly ordered by $\in$, then $(M,\in)$ is a well-order. In particular, it is order isomorphic to some ordinal.

On the other hand, if we drop the axiom of foundation from ZFC, it is consistent that there is a set $M$ which is linearly ordered by $\in$ and order isomorphic to $(\mathbb{R},<)$. Aczel's anti-foundation axiom (which is relatively consistent with ZFC without foundation) asserts roughly that the membership relation on the transitive closure of some set can look like any directed graph. We can apply this to the directed graph $(\mathbb{R},<)$, thinking of the instances of the order relation $<$ as directed edges between real numbers.

Title: Is there a model exists, such that homomorphic to the set of real numbers but the order is given by $\in$?

Body:
The question is motivated by the construction of a natural number system, where its order is given by $\in$, so is there any analogous exists for the set of real numbers?

It seems impossible for me since the cardinality of the set of real numbers is $2^{\aleph_0}$, but I am not really sure about this. To make a question precise, is there any model (with some order) $\mathcal{M}$ such that $\mathcal{M}\models T$ where $T$ is the theory of real numbers, and the order on $\mathcal{M}$ is given by $\in$?

Sorry if I am writing with ignorance with mathematical logic, I know very little about the topic.


问:
Help solving an Elliptic Integral
I am facing an elliptic integral to solve during thesis, however, I'm having a lot of trouble trying to solve it. The integral is given by:
$$\int{\frac{\sin^2(\theta)}{\sqrt{1+B\sin^2(\theta)}}}d\theta$$
where $B$ is a constant. I've been trying to use the Byrd and Friedman's Handbook but could not figure out some transformations that he adopts throughout the book. Any help provided will be greatly appreciated.

答:
$$\frac{\sin^2(\theta)}{\sqrt{1+B\sin^2(\theta)}}=\frac{\sqrt{B \sin ^2(\theta )+1}}{B}-\frac{1}{B \sqrt{B \sin ^2(\theta )+1}}$$
$$\int{\frac{\sin^2(\theta)}{\sqrt{1+B\sin^2(\theta)}}}d\theta=\frac{E(\theta |-B)-F(\theta |-B)}{B}$$

Title: Help solving an Elliptic Integral


问:
How to find the angle shown in figure?
The triangle shown in figure

I have to calculate the angle $\theta.$ I calculate the angles $135^o$ and $15^o.$ Also tried to use law of sine. but don't know how to find the angle $\theta.$

答:
I'm sure there'a a better and faster way, but this is how I'd solve it

Describe the oblique sides of the left triangle by the straight lines on which they sit, (and let the length of the horizontal side be $1$) i.e., $y=\frac{x}{\sqrt{3}}$ and $y=x-1$ (because $\tan 30º=\frac{1}{\sqrt{3}}$ and $\tan 45º=1$). The vertices of this triangle are located at $(0,0),(1,0)$ and $(\frac{1}{2}\left(3+\sqrt{3}\right),\frac{1}{2}\left(1+\sqrt{3}\right)).$

The rightmost side of the triangle to the right lies on the same line as $(2,0)$ and $(\frac{1}{2}\left(3+\sqrt{3}\right),\frac{1}{2}\left(1+\sqrt{3}\right)).$ This line has slope $2+\sqrt{3}$ and corresponds to an angle of $75º$. Therefore $\theta=105º$.

Title: How to find the angle shown in figure?


问:
Finding $n$ such that $(3-3w+2w^2)^{4n+3}+(2+3w-3w^2)^{4n+3}+(-3+2w+3w^2)^{4n+3}=0$ for $w\neq1$ a cube root of unity

Let
$${\left( {3 - 3\omega  + 2{\omega ^2}} \right)^{4n + 3}} + {\left( {2 + 3\omega  - 3{\omega ^2}} \right)^{4n + 3}} + {\left( { - 3 + 2\omega  + 3{\omega ^2}} \right)^{4n + 3}}=0$$
If $\omega\ne1$ be the complex cube root of unity, then the possible values of $n$ is

(A) $1\quad$ (B) $2\quad$ (C) $3\quad$ (D) 4

My approach $1+\omega+\omega^2=0$

Let $A={\left( {3 - 3\omega  + 2{\omega ^2}} \right)}$

$B={\left( {2 + 3\omega  - 3{\omega ^2}} \right)} $

$C= {\left( { - 3 + 2\omega  + 3{\omega ^2}} \right)}$

$B=A\omega$, $C=A\omega^2$ as $A+B+C=0$. Not able to proceed

答:
Your sum is $A^{4n+3}(1+\omega^{4n+3}+\omega^{2(4n+3)})$, and $A\ne 0$. So we must have $1+\omega^{4n+3}+\omega^{2(4n+3)}=0$. And $\omega^3=1$. So this is
$$1+\omega^n+\omega^{2n}=0$$
which is true if and only if $n\equiv 1$ or $2\bmod 3$.

As to which of the four answers to select, that is a different puzzle...


问:
Conditional expectation for the smallest sigma-field
It is claimed that for $\mathcal{G}=\{0,\Omega\}$ (smalles $\sigma$-field) the $\int_GE[X|\mathcal{G}]dP=\int_GXdP$ for $G\equiv\Omega$ implies that $E[X|\mathcal{G}]=E[X] a.e.$

I am having trouble understanding why this is so. I understood for the largest $\sigma$-field based on https://stats.stackexchange.com/questions/319941/int-gex-mathcalgdp-int-gxdp-implies-that-ex-mathcalg-x, but I don't get how to comprehend the smallest $\sigma$-field.

Any help would be appreciated!

答:
The random variable $Y:=\mathbb E[X\mid\mathcal G]$ is $\mathcal G$-measurable, which implies that for all $t\in\mathbb R$, the set $A_t:=\{Y\leqslant t\}$ is either the emptyset or $\Omega$. Letting $t_0:=\sup\{t\mid A_t=\emptyset\}$, we can see that $Y$ is constant, equal to $t_0$. Now, in order to determine the value of the constant, we use the definition of conditional expectation
$$
\int_GE[X|\mathcal{G}]dP=\int_GXdP
$$ with $G=\Omega$. The left hand side is $t_0$ and the right hand side $\mathbb E[X]$.


问:
Find the radius of the circle under these given conditions.

I tried to solve the question by considering the circle to be an incircle of 2 triangles joined as in this manner :

Now, this is my solution :

Let AE =X then AF =x ( properties of tangent )
This means , EB = 5 -x
Implies BC =5-x
By doing this to the lower triangle, we get CD also as 5-x

BC+CD = BD = 6
Then , we find X. How do we calculate r after this? Have I gone wrong in my method ?

答:
Alternative way:

Join the center of the circle $O$ to $E$. Then $$AO= AC-OC = 4-r$$ and $$OE=r,$$ so $$\sin \angle BAC = \frac 35 = \frac{OE}{AO} =\frac{r}{4-r} \implies r=\frac 32. $$ Note here I am considering the diagram you drew to refer to points.


问:
Find one fake coin out of 14 in 3 scales
There is a pile with 14 coins. One of them is fake and has a different weight than the others, which are the same.

Using a balance scale, find the fake coin with no more than 3 scales.

Based on the wiki article it is believed that no more than 13 coins can be tested with 3 scales, but I found a solution for 14 coins, and I wish to share it with you in my answer below.

答:
Take two coins and slice them in half with equal weight.

Scale 1: $3+0.5 + 0.5$ vs. $4$ -> Take 2 distinct half coins with 3 other coins, let's call it "pile A", and weigh them against 4 other coins. Let's call it "pile B". And we are leaving 5 coins on the side.

If the scale is not balanced, it means that the 5 coins we didn't scale are not fake.

Scale 2: $2 + 3B$ vs. $1 + 4R$ -> Take 2 coins from pile A with 3 coins from pile B and weight them against 1 coin from pile B and 4 real coins(those we didn't weight and we now know are not fake)

There are three possible outcomes: the scale is flipped, the scale is balanced, or it didn't change. In all those cases, we left out with three coins where we know the relative weight of a fake coin.

Scale 3: $1$ vs. $1$: Weight two of the coins in question, one against the other, leaving one aside. In the case of half coins, compare half and half coins. If the scale is balanced, then the fake coin is the one we put aside. Otherwise, we should know who is the fake coin from the scale result, as in this case, we know the relative weight of the fake coin.

In case the scale is balanced, it means that the fake coin is among the 5 coins we didn't weight

Scale 2: $2$ vs. $1 + 1R$ -> Take two coins, let's call it "pile A" and scale them against 1 coin and 1 real coin, let's call it "pile B".

If the scale is balanced, then the fake coin is one of the 2 we didn't scale. Scale one of them with a real coin to find out who is the fake coin.

Scale 3: $1 + 1B$ vs. $2R$ -> If the scale is not balanced, then scale one coin from pile A with one coin from pile B against two real coins. There could be three outcomes: The scale is balanced; the scale is flipped; the scale didn't change. That correlates with the 3 coins we have been testing to find the fake coin.


问:
Definition of metastability
I was reading Terence Tao's blog post on analogies between soft and hard analysis when I saw that the soft analysis statement "$x_n$ is convergent" corresponds to the hard analysis statement "$x_n$ is metastable".

What does "metastable" mean? The Wikipedia article on that term explains it only in the context of physical systems:

Metastability describes the behaviour of certain physical systems that can exist in long lived states that are less stable than the system's most stable state.

I understand that a sequence $x_n$ can be viewed as a (discrete-time) dynamical system, but I need a precise mathematical definition of this property that sequences can have.

答:
See this paper by Avigad/Dean/Rute or this other paper by them. I'm not sure (haven't looked very hard into the matter), but the term metastable in this context may be something originally coined by Tao, perhaps beginning with this paper by Tao, although your latest comment suggests the term originated with someone else.

I glanced at Tao's blog on soft and hard analysis, but I'd have to spend more time than I have at the moment (or even the next few days, due to upcoming deadlines at work) to get a good sense of what metastable convergence means. [Side note: Off the top of my head, I know a couple of other examples of soft/hard analysis examples that are not mentioned in Tao's blog. One is E. H. Moore's notion of relatively uniform convergence for a series of functions, and another is the classification of nowhere dense sets by generalized notions of porosity (such as I observe in the middle of p. 574 of this paper).] A possibly good place to look for pointers to expository items (conference slides, class notes, etc.) about what metastability of sequences means is Jeremy Avigad's homepage, but for some reason I can't get his webpage to load right now.

Maybe by posting this (as you suggested in a comment) someone who knows more than I do will chime in, as my answer will bump your question back to the top of the questions for others to see.

Title: Definition of metastability


问:
Closed form of a sum with a falling factorial
I am having trouble finding the closed form for this:

$\sum_{k=1}^n  k^\underline{-2} + 2^k$

If it were just

$\sum_{k=1}^n 2^k + 2k$

I would just use the geometric series to solve for the closed form, but I do not know how to apply the geometric sum to the falling factorial. Can someone help me out?

答:
Using the definition of Falling Factorial, rewrite the sum as
$$\begin{align} \sum_{k=1}^n  k^\underline{-2} + 2^k &= \sum_{k=1}^n k(k-1) + 2^k \\&= \sum_{k=1}^n k^2-k + 2^k \\&= \sum_{k=1}^nk^2 - \sum_{k=1}^nk + \sum_{k=1}^n 2^k \end{align}$$ Then, use the known formulas.


问:
Where does p-value should be drawn in case of two tailed Chi squared test?
Here is an illustrative example of one tailed test:

In case of one tailed test, the p-value would be area to the left of the critical value 57.336 but in case of two tailed test the p-value is doubled. How to know where to draw the second area?

答:
Your first chart illustrates the two-tailed situation.  You are trying to have the two tails having the same probability so you can say they cover values as extreme as or more extreme than the observation.  You can answer your question using the cumulative distribution function and its complement and inverse.

Let's take an example, where you observe the value of $65$ in your $\chi^2_{39}$-distribution; this above the median (near $39$).

R gives pchisq(65,39) as $\mathbb P(X \le 65) \approx 0.994416$.

Subtract form $1$ to get $0.005584$ as the probability of being extreme in a one-tailed sense

Double this to get the two-tailed $p$-value of $0.011168$

The other tail corresponds to the distribution below qchisq(0.005584,39) about $20.21$

You can get this second value directly with qchisq(1-pchisq(65,39),39)

Incidentally, $\chi^2$ tests are typically one-tailed unless you are investigating fraud, as low $\chi^2$ statistics are typically associated with sample distributions close to the null hypothesis's prediction and so might not be seen as extreme.


问:
Empty interval or not?
Looking for my problem I've found an answer about completely empty interval:
https://math.stackexchange.com/a/1228313 that's clear.

But what about this thing: [0,0) or [x,x) - an inclusive beginning and an exclusive end?
Does it make x a part of this interval, or does the "exclusiveness" take it out? What is then taken as "higher", inclusiveness or exclusiveness? Does it make sense whatsoever?

答:
It is an empty interval, not even a point.
for an interval (a,b) or [a,b) or (a,b], b>a should be required.
but for [a,a], it could be considered as a point with no length.</p>

Title: Empty interval or not?


答:
Your intuition was right. Here is the formula and the correct solution:

$r=\frac{Area}{half-perimeter}=\frac{6\times 4:2}{(5+5+6):2}=\frac{12}{8}=\frac{3}{2}$

Title: Find the radius of the circle under these given conditions.

Body:
I tried to solve the question by considering the circle to be an incircle of 2 triangles joined as in this manner :

Now, this is my solution :

Let AE =X then AF =x ( properties of tangent )
This means , EB = 5 -x
Implies BC =5-x
By doing this to the lower triangle, we get CD also as 5-x

BC+CD = BD = 6
Then , we find X. How do we calculate r after this? Have I gone wrong in my method ?


问:
Show $\mathcal{R}$ related
If $\alpha\in T_n$ and $\beta\in S_n$ show that $\alpha \mathcal{R} \alpha\beta$ ($T_n$ is the full transformation monoid, and $S_n$ is the symmetric group, both on $\{1,2,\ldots ,n\}$).

Does this mean show $\alpha T_n = (\alpha\beta )S_n$? How would you do it?

答:
Would $\alpha T_n = (\alpha\beta )S_n$ then $\alpha T_n = \alpha S_n$ what  generally speaking is not true.

Since $\alpha\beta \in \alpha T_n\beta \subset \alpha T_n$ and  $\alpha \in \alpha\beta T_n\beta^{-1} \subset \alpha\beta T_n$ then $\alpha \mathcal{R} \alpha\beta$.

Title: Show $\mathcal{R}$ related


问:
$\langle x,x_n\rangle \leq \|x_n\|$
I am wondering if we have that $x_n\to x$ weakly, is it then true that $\langle x,x_n\rangle \leq \|x_n\|$, and why?

答:
For me it is wrong.

On $\mathbb{R}$ take $x=2$ and $x_n = 2- \frac{1}{n}$.

Then $\langle x,x_n\rangle = 2\left(2- \frac{1}{n}\right)$, $\| x_n\|=2- \frac{1}{n} < 2\left(2- \frac{1}{n}\right)$

and of course $x_n \to x$ strongly so a fortiori weakly.

Title: $\langle x,x_n\rangle \leq \|x_n\|$


问:
Solving system of equations using Chinese remainder theorem
I have a system of equations on that I would like to solve programatically:

$(x + a_1) \bmod b_1 = 0$

$(x + a_2) \bmod b_2 = 0$

$...$

$(x + a_i) \bmod b_i = 0$

$a$ and $b$ are given, and I would like to find the smallest positive solution for $x$. The numbers are too big to iterate through all possible solutions. I think it should be possible to solve it using the Chinese  remainder theorem, but I haven't been able to get from one to the other. How can I translate this problem into a problem that can be solved using the CRT?

答: Your system of equations is the same as solving $x\equiv -a_i \pmod{b_i}$.


问:
Paradigm Shifts in Mathematics
in physics there were several clear revolutions or paradigm shifts which fundamentally changed the field. One example is the copernican revolution and the encompassing shift from the ptolemaic to heliocentric view.

Given that mathematics works from axioms, I figured it is unlikely for wrong assumptions to creep into the canon of the field.
Additionally, during my math education (as a physicist) I had the feeling, that mathematics evolved rather continuously from the greeks to today, always adding new knowledge on top of the old.

Therefore my question is, whether this is wrong and there were certain paradigm shifts or radical reinterpretations of previous results in the history of mathematics, or was it a continuous growth of knowledge?

Addendum

There has been this question already, which is asking for philosophical shifts  in mathematics. However, I figured it is different from this one, since I try to understand whether the body of mathematical knowledge grew linearly or was discontinuous at certain points.

答:
I suppose we might distinguish "revolutions" which bury their dead (so to speak) from "paradigm shifts" (where the game moves on, and work done in the old style is not expunged but no longer looks interesting or important to pursue).

I suppose it was once thought that the 19th century reworking of analysis without infinitesimals was a revolution that displaced falsehood/incoherence (which is why varieties of non-standard analysis which rehabilitated infinitesimals -- sort of! -- came as an intriguing surprise a hundred and something years later). The development of set theory was a revolution, in showing that it was possible to have a coherent theory (of "completed infinities") where previously it was thought that there could only be falsehood/incoherence.

But these sort of cases are surely the exception (in maths at any rate). A paradigm shift need not involve supposing that what has gone before is wrong.  Rather,  novel concepts are introduced, new problems can be  raised, new approaches come to be seen as particularly interesting/rewarding; new exemplars come to be regarded as paradigms to be emulated, and as setting the standards by which problem-solutions are judged. The development of abstract algebra in the last century, for example, would seem to be  a paradigm example of this sort of paradigm shift ...!

Title: Paradigm Shifts in Mathematics


问:
Not aligning denominator ahead of time breaks the equation, why?
Why can I not align the denominator later on? (it gives me completely different values)

I thought I already got this down pat, but I guess not. Can someone point me to where my knowledge gap is, and where I need to brush up?

答: They are actually the same! When you expand the top expression, you should get the bottom one. Note the bottom expression is under the category Alternate forms, so they are all the same, just written in different forms.


问:
Covering an 8x8 board with L and O Tetromino
I solved a puzzle about proving that if a rectangular board can be covered by L-Tetrominoes then the number of squares must be a multiple of 8.

I based the solution on a colored board (like a chessboard) and 2 types of L-Tetrominoes

Either:

```<code> _____
|BLACK|
|WHITE|_______
|BLACK| WHITE|
--------------
</code>```

or

```<code> _____
|WHITE|
|BLACK|_______
|WHITE|BLACK |
--------------
</code>```

The follow up puzzle is:

Prove that an 8x8 board can not be covered with 1 O-Tetromino and 15 L-Tetrominoes.

The O-Tetromino is:

```<code> ____ ____
|    |    |
|----+----|
|____|____|
</code>```

I thought of the following proof:

If I use 1 O-Tetromino then there remains a board of 8x8-4 = 60 squares.

60 is not a multiple of 8 and we have already shown that for a board to be covered with L-Tetrominoes must be a multiple of 8. Therefore the 8x8 board can not be covered with 1 O-Tetromino and 15 L-Tetrominoes

Is this proof valid? It seems to informal also in relation to the approach I followed to prove that the L-Tetrominoes cover a board of x8

If it is not, then how can I prove it?

答:
If you have stated the earlier conclusion correctly, the proof does not work because the board less the O tetromino is not rectangular.

A simpler proof that you need an even number of tetrominos to cover a rectangle comes from coloring the board with alternate black and white stripes instead of the traditional checkerboard.  With that coloring, each L tetromino covers three squares of one color and one square of the other.  An odd number of L tetrominos cannot color an equal number of black and white squares.  Now once you note that the O tetromino covers two squares of each coloring, you can just use the count of squares instead of the shape of the board to conclude that $15$ L tetrominos cannot cover what is left.


问:
How to prove $\mathbb{Z}[i]$ is a ring but not a field?
Let i be the imaginary unit of $\Bbb{C}$ and $\mathbb{Z}[i]:= \{a + ib, ab \in\mathbb{Z} \} $

So the task is to prove that this a ring regarding multiplication and addition in $\mathbb{C}$ which is commutative, free of zero divisors and posseses a 1 element.
I already found out that it is a ring and that the 1 Element is (1,0) and 0 Element is (0,0), so in both cases $i= 0$. So I guess this is like proving a trivial Ring.
I am just confused because I am not sure about the $i$. How would I complete the proof? I am stuck because I don't know how to prove that it is free of zero divisors and not a field.
Thank you for the help.

答:
The quotient field of $\Bbb Z[i]$ obviously is $\Bbb Q(i)$, which is different from $\Bbb Z[i]$. Hence  $\Bbb Z[i]$ cannot be a field, because the quotient field of an integral domain $R$ also is the smallest field containing $R$.

References with many details:

Quotient field of gaussian integers

Prove the fractional field of an integral domain is the smallest field containing the integral domain


问:
What is the chance that an infinite series of random numbers will, at some point, repeat all previously generated numbers in that order?
The main reason of this post is to provide a writeup of what I think is a correct solution to this problem from the r/math subreddit. I'll quote the poster to phrase the problem:

What is the chance that an infinite series of random numbers will, at some point, repeat all previously generated numbers in that order?

While there was some confusion in the comments, I believe what the poster means is something like the sequences below:
$$7,7,...$$
$$3,2,3,2,...$$
$$7,8,6,7,8,6,...$$
Where the dots afterward are arbitrary as the condition has already been fulfilled. If we start writing out a sequence of digits in this way, where the probability of selecting any digit is equal, what is the probability that something like this will happen?

I'll post my solution below. Feel free to comment or correct me.

答:
Your solution is hard to follow, even the definitions.  Not sure why you skip from using $e$ as an example to using $\pi$, nor why you appear to make an issue of the first decimal in $e$.  A string of digits doesn't have a natural location for the decimal point.  Also, why the special interest in powers of $2$?

In any case, let's just consider the simpler problem of the first $6$ digits.

There is, of course, a $\frac 1{10}$ chance that the first two digits match. So $p_1=.1$  There are, of course, $10$ strings of length $2$ of that form.

Failing that, we consider strings of the form $XYXY$ where $X\neq Y$.  There are $90$ strings of that form, out of a possible $10^4$.  So $p_2=\frac {90}{10000}=.009$

For $p_3$, there are $10^6$ strings of length $6$ of which $10^3$ have the same first $3$ and last $3$ digits.  Of these, only those where the first and third digits match replicate after $4$, and only those where the first and seconds digit match replicate after $2$.  So we just need the first digit to be distinct from the second two.  There are, therefore, $10\times 9^2$ good strings of length $6$, so $$p_3=\frac {10\times 9^2}{10^6}=.00081$$

But then $$p_1+p_2+p_3=.10981$$  which is already greater than your answer.


问:
A problem on Principle of inclusion and exclusion.

I am having a little bit of trouble understanding this part in my book. I know the normal PIE that is n(A+B) = n(A) + n(B) - n(AB) and then so on for more sets but I am not able to understand this form.  What do they mean by numerical value associated is constant, what is meant by N(i) common value?

So could someone please explain this to me.

Thank you!

答:
They are inventing notation to simplify things... in the case that $n(a_1a_2)=n(a_1a_3)=n(a_1a_4)=\dots=n(a_2a_3)=\dots=n(a_ia_j)$ for all $i<j$ then rather than writing $n(a_1a_2)$ they write $N(2)$.  Similarly, in the case that $n(a_1a_2a_3)=n(a_1a_2a_4)=\dots=n(a_ia_ja_k)$ for all $i<j<k$ then instead of writing $n(a_1a_2a_3)$ they just write $N(3)$ and so on for larger tuples.

Reworded, they are saying that in the event that all pairs have the same size as each other, all triples have the same size as each other, etc... then:

$|a_1\cup a_2\cup \dots \cup a_n|=n\cdot |a_1|-\binom{n}{2}\cdot |a_1\cap a_2| + \binom{n}{3}\cdot |a_1\cap a_2\cap a_3|-\dots + (-1)^{k+1} \binom{n}{k}|a_1\cap a_2\cap \cdots \cap a_k|+\dots$

They just wanted to shorten the notation, rather than saying "The size of the intersection of the $k$ sets $|a_1\cap a_2\cap \dots \cap a_k|$" they instead just say "The size of the intersection of $k$ sets $N(k)$."


问:
Line Integral of Every Positively Oriented Simple Closed Path - Green's Theorem
This question is from Example #5, Section 16.4 on P1059 of Calculus, 6th Ed, by James Stewart.

Given Question: If $\mathbf{F}(x,y) = \left(\dfrac{-y}{x^2 + y^2}, \dfrac{x}{x^2 + y^2}\right)$, show that $\int_C \mathbf{F} \cdot d\mathbf{r} = 2\pi $ for every positively oriented simple closed path that encloses the origin.

Part of the Given Solution: Since $C$ is an ARBITRARY closed path that encloses the origin, it's difficult to compute the given integral directly. So let's consider a counterclockwise circle $A$ with center the origin and radius $a$, where $a$ is chosen to be small enough that $A$ lies inside $C$, as indicated by the picture below. Let $D$ be the region bounded by $C$ and $A$.

  Then positively oriented $\partial D = C \cup (-A)$. So the version of Green Theorem's applied to regions with holes gives:

$\int_C \mathbf{F} \cdot d\mathbf{r} + \int_{-A} \mathbf{F} \cdot d\mathbf{r} = \iint_D \underbrace{(\partial_x Q - \partial_y P)}_{\Large{= 0}} \, dA $
  $\Longrightarrow \int_C \mathbf{F} \cdot d\mathbf{r} = \int_{A} \mathbf{F} \cdot d\mathbf{r}$.

  (Rest of solution omitted)

$\Large{\text{Q1.}}$ I can't perceive how one would divine to construct $A$ to solve this problem. So I feel that my following solution is more intuitive. Is it really? However, I don't seem to need Green's Thoerem. How and where would it be applied?

My solution:  It's hard to calculate $\int_C \mathbf{F} \cdot d\mathbf{r} $. The idea then is to find one $A$ so that

$(\dagger) \int_{A} \mathbf{F} \cdot d\mathbf{r}$ is easier to compute and

$(\ddagger)\int_C \mathbf{F} \cdot d\mathbf{r} = \int_{A} \mathbf{F} \cdot d\mathbf{r}$.

Now, $(\ddagger) \iff \int_C \mathbf{F} \cdot d\mathbf{r} \huge{\color{red}{-}} \normalsize\int_A \mathbf{F} \cdot d\mathbf{r} = 0 \iff \int_C \mathbf{F} \cdot d\mathbf{r}  +\int_{\huge{\color{red}{-}\normalsize{A}}} \mathbf{F} \cdot d\mathbf{r} = 0 $
$\iff \int_{\Large{\partial D} \normalsize{ \, = \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})}} \mathbf{F} \cdot d\mathbf{r} = 0$.

For $\partial D$ to be positively oriented, $C$ and $\color{red}{-}A$ must both be positively oriented $\Longrightarrow A$ must be NEGATIVELY oriented, so $D$ must be to the right of $A$. One convenient choice of $A$ would just be a counterclockwise circle with center the origin and the radius $a$.

$\Large{\text{Q2.}}$ Why don't I get the requested answer if I pick $A$ as a counterclockwise ellipse enclosing the origin? In other words, $A$ is parameterised via $r(t) = (a\cos t, b\sin t), 0 \leq t \leq 2\pi $.

Then $\int_{A} \mathbf{F} \cdot d\mathbf{r} = \int_0^{2 \pi} \left(\dfrac{-b\sin t}{a^2 \cos^2 t + b^2 \sin^2 t},\dfrac{a\cos t}{a^2 \cos^2 t + b^2 \sin^2 t} \right) \cdot (-a\sin t, b \cos t) \, dt$.

$ = ab \int_0^{2 \pi} \dfrac{1}{a^2 \cos^2 t + b^2 \sin^2 t} \, dt $.

If $a = b$, then the requested answer follows by inspection. But what if $a \neq b$?

$\Large{\text{Supplementaries to Muphrid's Answer: }}$

$\Large{\text{Q1.1.}}$ To rephrase your answer, are you saying that my solution has to justify why there exists an $A \ni \int_C \mathbf{F} \cdot d\mathbf{r} = \int_{A} \mathbf{F} \cdot d\mathbf{r}. ?$
This is true because $\int_C \mathbf{F} \cdot d\mathbf{r} = \int_{A} \mathbf{F} \cdot d\mathbf{r} \iff \int_{\Large{\partial D} \normalsize{ \, = \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})}} \mathbf{F} \cdot d\mathbf{r} = 0, $
where the last equation is true by Green's Theorem: $\int_{\Large{\partial D} \normalsize{ \, = \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})}} \mathbf{F} \cdot d\mathbf{r} = \iint_D \underbrace{(\partial_x Q - \partial_y P)}_{\Large{= 0}} \, dA. $

$\Large{\text{Q1.2.}}$ I wrote that $\partial D \normalsize{ \, = \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})} $. Are you saying that this is wrong and that $\partial D \normalsize{ \, \subsetneqq \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})} $?

$\Large{\text{Q1.3.}}$ I understand that $\mathbf{F}(0,0)$ is undefined. However, how does this imply that integrals enclosing the origin pick up $2\pi$?

$\Large{\text{Q1.4.}}$ How is $(\nabla \times \mathbf{F}) \cdot d\mathbf{A} = \partial_x Q - \partial_y P $? How is $d\mathbf{A} = z \, dx \, dy ?$

$\Large{\text{Q2.1.}}$ I understand that using a circle works fine and is easier. I just want to derive the same answer with an ellipse though. What went wrong?

$\Large{\text{Supplementaries to Muphrid's 2nd Comment: }}$

$\Large{\text{Q1.3.1.}}$ You write: "When you choose a unit circular path, the resulting integral looks like an arclength integral."

Are you saying that choice of a unit circular path $\Longrightarrow \int_{A} \mathbf{F} \cdot d\mathbf{r} = ... = \int_{0}^{2\pi} 1 \, dt = 2\pi? $ How are you so confident that $...$ always yields this answer?

$\Large{\text{Q1.4.1.}}$ Why can you define $d\mathbf{A} = z \, dx \, dy ?$ We're not given $d\mathbf{A}$?

答:
1) First, while you've chosen the orientation of $A$ correctly, you should realize that the boundary $\partial D$ of the region $D$ contains $C$ and $-A$.  This is important.

Now, apply Green's theorem:

$$\oint_{\partial D} F \cdot dr  = \int_C F \cdot dr - \int_A F \cdot dr = \int_D (\nabla \times F) \cdot dA$$

You should argue that the area integral is zero.  Instead, in your second point ($\ddagger$), you come to this conclusion but without the supporting logic to back it up (which must come from Green's theorem).

(For fun: realize that $\nabla \times F \neq 0$ at the origin, and this is why integrals enclosing the origin pick up $2\pi$.)

2) Why would you pick an ellipse instead of a circle?  Green's theorem tells us that $C$ could be an ellipse instead of an arbitrary curve and thus the integral would have to be $2\pi$, but why put yourself through that work?

Responses to supplemental questions:

1.1) Yes, that is the correct logic.

1.2) No, that is the correct expression for the boundary.

1.3) I'm just observing that the line integral enclosing the origin can only be nonzero if the curl is somewhere nonzero in the region enclosed.  You can also understand this in relation to complex analysis, with the point at the origin being a pole, which has a nonzero residue.

1.4) Yes, I wrote this incorrectly.  I've added the necessary cdot to the original formula.

2.1) It doesn't seem to me there's anything wrong with that integral; it's just not easy to solve (wolfram gives the right answer for specific values of $a$, $b$, but not symbolically).  You could try, for example, the polar equation of an ellipse, but I can't guarantee that would be any easier.

Responses to supplemental supplemental questions:

1.3.1) This problem is very specifically constructed so that $F = \hat \theta /r$.  Note the hat: this denotes a unit vector.  $\hat \theta$ is always tangent to curves of constant $r$ and is in the direction of increasing $\theta$.  When you choose a unit circular path, $r = 1$, and $d\mathbf r/d\theta = r \hat \theta$ also, rendering the integral

$$\int_A F \cdot d\boldsymbol \ell = \int_0^{2\pi} \frac{\hat\theta}{r} \cdot \frac{\hat \theta}{r} \, d\theta = \int_0^{2\pi} \hat \theta \cdot \hat \theta \, d\theta$$

Since $\hat \theta \cdot \hat \theta = 1$, the form I argued follows.  That is precisely why choosing a unit circular path is so convenient here.

1.4.1) You have the freedom to choose the surface whose boundary is a given curve.  That choice of surface then determines the differential. I don't boldface vectors, so you'll have to figure out from context whether a quantity is scalar or vector.  Nevertheless, I wrote $dA = \hat z \, dx \, dy$, and $\hat z$ is a unit vector.  The choice of surface that goes with this is that of using the flat $xy$-plane, which is the most convenient (I dare say necessary, as you have no information about what $F$ does outside of the $xy$-plane).

Edit: the definition of $\hat \theta$ is

$$\hat \theta = \frac{1}{r} \frac{\partial \mathbf r}{\partial \theta}$$

write $\mathbf r = (r \cos \theta, r \sin \theta)$, and you get

$$\hat \theta = \frac{1}{r} (-r \sin \theta, r \cos \theta) = \left(- \frac{y}{\sqrt{x^2 + y^2}}, \frac{x}{\sqrt{x^2+ y^2}} \right)$$

You should be able to see that $\hat \theta/r = F$ then.


答:
Mathematics is not an axiomatic discipline. One way a new field is opened is by generally by uncovering examples which have something in common and which appears to point to a new theory.

Take for example homology. This was axiomatised by Eilenberg & Steenrod. But had not people discovered the Betti numbers, had not Poincare discovered homology and had not Noether pointed out that the Betti numbers wrre better thought of as groups there would not have been something to axiomatise.

Hilbert says more or less the same in his Geometry & the Imagination where he classes deductive thinking, that is the thinkng that comes from the axiomatic form of a lower order than that of inductive thinking which he classifies as the true form of scientific thinking.

Personally, a key paradigm shift for me has been the introduction of category-theoretic thinking into mathematics and it also demonstrates the continuity of thought too. For example, the triangle was discovered early, by adding directions to the sides we have the law of vector addition and then by allowing the sides to be curved we can think of them as category-theoretic arrows. This is also revealing:  we can think of them as non-Euclidean vectors and in a length space where between any two points there is a unique geodesic we can lift the directed geodesics into such a vector.

Title: Paradigm Shifts in Mathematics

Body:
in physics there were several clear revolutions or paradigm shifts which fundamentally changed the field. One example is the copernican revolution and the encompassing shift from the ptolemaic to heliocentric view.

Given that mathematics works from axioms, I figured it is unlikely for wrong assumptions to creep into the canon of the field.
Additionally, during my math education (as a physicist) I had the feeling, that mathematics evolved rather continuously from the greeks to today, always adding new knowledge on top of the old.

Therefore my question is, whether this is wrong and there were certain paradigm shifts or radical reinterpretations of previous results in the history of mathematics, or was it a continuous growth of knowledge?

Addendum

There has been this question already, which is asking for philosophical shifts  in mathematics. However, I figured it is different from this one, since I try to understand whether the body of mathematical knowledge grew linearly or was discontinuous at certain points.


问:
How to find the angle shown in figure?
The triangle shown in figure

I have to calculate the angle $\theta.$ I calculate the angles $135^o$ and $15^o.$ Also tried to use law of sine. but don't know how to find the angle $\theta.$

答:
I managed to obtain a synthetic solution. This is my first time posting a solution, so let me now if I do something wrong.

The idea I followed is to create isosceles and equilateral triangles until I used all the information given in the question. I first split the 135 degree angle into 120 and 15 degrees. In the newly created quadrilateral, I connected the diagonals. Finally, I marked equal sides until obtaining the answer.

Title: How to find the angle shown in figure?


问:
The Fermat-Catalan conjecture with exponents $(2,n,4)$, $n\ge4$
The Fermat-Catalan conjecture is that for coprime $x,y,z$ and positive integers $a,b,c$ with $1/a+1/b+1/c<1$, the generalized Fermat equation $x^a + y^b = z^c$ has only finitely many solutions. I'm considering exponent triples $(a,b,c)$ which are solved.

Table 1 of [BCDY] surveys known results and states that $(2,n,4)$, $n\ge4$ has been solved completely and that this is 'Immediate from Bennett–Skinner [BS], Bruin [Br3]'. [Br3] covers the case $n=5$. Fermat dealt with $n=4$.

This leaves $n=6, 9$ and prime $n\ge7$, but I can't see how [BS] is relevant to that. Can someone explain and/or point me to the relevant part of [BS].

[BCDY] 'Generalized Fermat equations: A miscellany', Bennett, Chen, Dahmen, Yazdani, International Journal of Number Theory, Vol. 11, No. 1 (2015)

[BS] 'Ternary Diophantine Equations via Galois Representations and Modular Forms', Bennett, Skinner, Canad. J. Math. Vol. 56(1), 2004 p23-54.

[Br3] 'Chabauty methods using elliptic curves', Bruin, J.reine angew. Math. 562 (2003), 27-49.

答:
I reposted this question in MO, and have copied this answer from there.

[Br2] Theorem 1 covers the case $n=6$. So this leaves $n=9$ and prime $n\ge7$.

As suggested in a comment on MO, I contacted Michael Bennett directly and he kindly explained the rest to me:

We have $x^2+y^n=z^4$ with $x,y,z$ coprime integers.

So $(z^2 - x)(z^2 + x) = y^n$.
The gcd of $(z^2 - x)$ and $(z^2 + x)$ is $1$ or $2$.

For a gcd of $1$ we have $x$ and $z$ of opposite parity, and can write

$z^2-x = u^n$ and $z^2+x = v^n$

so that $u^n+v^n = 2z^2$.

This is solved for coprime integer $u,v,z$ for $n\ge4$ by [BS] Theorem 1.1.

For a gcd of $2$ we have $x$ and $z$ both odd, and one of

$z^2-x = 2 u^n$ and $z^2+x = 2^{n-1}v^n$, or

$z^2+x = 2 u^n$ and $z^2-x = 2^{n-1}v^n$.

In either case, $u^n + 2^{n-2} v^n = z^2$.

This is solved for coprime integer $u,v,z$ for prime $n\ge7$ by [BS] Theorem 1.2.

This leaves the case $n=9$ with $y$ even.

Going back to the original equation, we have
$x^2+y^9=z^4$.
[Co, Section 14.4.1] gives complete parametrizations of
$x^2+w^3=z^4$
in terms of $s$ and $t$. In our case, $w$ is an even cube and from this and the parity constraints on $s$ and $t$ given in [Co] it follows that there exist coprime integers $s$ and $t$ with
$s t (s^3 - 16 t^3) (s^3 + 2 t^3)$ a cube.

The factors on the left hand side are pairwise coprime. ($s^3-16t^3$ and $s^3+2t^3$ could possibly have a common factor of $3$, but if they do, the whole expression is divisible by $9$ but not by $27$, and hence is not a cube.)

Since the factors are pairwise coprime, $s^3+2t^3$ is a cube.

This corresponds to a rational point on the curve $A^3+2B^3$=1 which is isomorphic to the elliptic curve $Y^2=X^3-1728$ via standard transformations. The latter curve has rank $0$ (and only the rational points corresponding to the point at infinity and $(X,Y)=(12,0)$). Tracing these back to $A^3+2B^3=1$, we find that $(A,B)=(1,0)$ or $(A,B)=(-1,1)$.

These points lead to either $t=0$ (which gives $z=0$ in $x^2+y^9=z^4$) or to $st = -1$ (which does not make $s t (s^3 - 16 t^3) (s^3 + 2 t^3)$ equal to a cube).

[Br2] 'The Diophantine Equations $x^2 \pm y^4 = \pm z^6$ and $x^2 + y^8 = z^3$', Bruin, Compositio Mathematica 118: 305-321, 1999.

[Co]  'Number Theory Volume II: Analytic and Modern Tools', Henri Cohen


问:
How do I write the Partial Sum of this (if it exists)?
I have a summation (really a whole limit): \begin{equation}\sum_{k=1}^\infty \frac{1}{n^2+k^2} \end{equation}
I have thought of the problem by writing out the problem as follows:
\begin{array}{|c|c|}\hline S_1&\frac{1}{n^2+1^2}\\ \hline S_2& \frac{1}{n^2+1}+\frac{1}{n^2+4}\\ ...& ... \end{array}
My thought progression is the numeration becomes a polynomial and the bottom another polynomial of infinite size. However, I also know that when I take the limit of the partial sum I get the value that converges as follows: \begin{equation} \lim_\limits{n\to \infty}S_n=L \end{equation}
Any thoughts to how I can get $S_n$ because if there is a  pattern I am not seeing it.
##Edit
This is really the original problem:
\begin{equation}\lim_{x\to\infty}x\lim_{n\to\infty}n\sum_{k=1}^\infty(\frac{1}{n^2+k^2}) \end{equation}
I am assuming that this represents this: \begin{equation} \lim_{x\to\infty}\int_0^x \frac{1}{1+t^2}dt \end{equation}
After a couple of suggestions I tried rewriting the preceding as the following:
\begin{equation}\lim_{x\to\infty}\int_0^x(1-t^2+t^4-t^6+t^8-t^{10}+O(t^{12}))dt \end{equation}

答:
$$S_p=\sum_{k=1}^p\frac{1}{n^2+k^2}=\sum_{k=1}^p\frac{1}{(k+i n)(k-in)}=\frac i{2n}\sum_{k=1}^p\left(\frac 1{k+in}-\frac 1{k-in} \right)$$

Now, using generalized harmonic numbers and the infinite limit
$$S_p=\frac{-i n \left(H_{p-i n}-H_{i n+p}\right)+\pi  n \coth (\pi  n)-1}{2 n^2}$$ Using asymptotics
$$S_p=\frac{\pi  n \coth (\pi  n)-1}{2 n^2}-\frac{1}{p}+\frac{1}{2
   p^2}+O\left(\frac{1}{p^3}\right)$$


问:
Integration by parts with cdf
Is there are any simplification available to
$$\int_{-\infty}^t(t-x)dF(x)$$ where, $F(x)$ is a cdf.

I tried with integration by parts
$$\left[(t-x)F(x)+\int F(x)dx\right]_{-\infty}^t$$

not sure how to progress beyond this, will the first part at $x=-\infty$ be zero? as $F(-\infty)=0$

答:
The term $(t-x)F(x)$ is zero also at $x=t$. Therefore, the integral simplifies to
$$\int_{-\infty}^t(t-x)\,dF(x) = \int_{-\infty}^t F(x)\,dx $$
More generally, the integral
$$
\Phi(t)=\frac{1}{(n-1)!}\int_a^t (t-x)^{n-1} f(x)\,dx
$$
performs $n$  integrations of the function $f$: it produces $\Phi$ such that $\Phi^{(n)}=f$ and $\Phi(a)=\dots=\Phi^{(n-1)}(a)=0$.</p>

Title: Integration by parts with cdf


问:
How to prove a differential entropy is not scale invariant?
For example, $S(X)=-E_X(\log(f_X))=-\int_{-\infty}^{+\infty}f_X(x)\log(f_X(x))dx$
A transformation of X changes the result:$S(aX)=S(X)+\log|a|$ and more in general $S(g(X))=S(X)+\int_{-\infty}^{+\infty}f_X(x)\log(|\frac{\partial g}{\partial x}|)dx$.

答:
You have correctly shown that the differential entropy of a random variable is not scale invariant. I encourage you not to spend any more time trying to prove that it is invariant under this transformation.

If you want a scale invariant measure for the entropy of a continuous random variable you need to use a different related quantity, for example the relative entropy.

Here is a proof that the relative entropy for the random variables $P$ and $Q$ is invariant under coordinate transformations (and hence under scale transformations).

The definition for the relative entropy is $$D_{KL}(P||Q) := \int dx f_P(x) \log\left(\frac{f_P(x)}{f_Q(x)}\right).$$ Under a change variable from $x$ to $y$, the pdfs of $P$ and $Q$ change as $$f_P(y)dy = f_P(x)dx\qquad\qquad f_Q(y)dy=f_Q(x)dx.$$ (Note that in this notation, $f_P(y)$ is a different function from $f_P(x)$). Then changing variable from $x$ to $y$ in the integral, we see that $$D_{KL}(P||Q) = \int dx f_P(x) \log\left(\frac{f_P(x)}{f_Q(x)}\right) = \int dy f_P(y) \log\left(\frac{f_P(y)\left|\frac{dy}{dx}\right|}{f_Q(y)\left|\frac{dy}{dx}\right|}\right).$$Hence we arrive at$$D_{KL}(P||Q) =\int dy f_P(y) \log\left(\frac{f_P(y)}{f_Q(y)}\right),$$showing invariance under coordinate transformation as required.


问:
Having problems this proof
Suppose that $\phi$ proves if $\alpha$ then $\neg\beta$ and that $\phi$ proves $\beta$. Can we infer anything from $\phi$?

答:
Yes, use DT and MP.

From $\Gamma \vdash \beta \to \lnot \alpha$ we have, using MP: $\Gamma \cup \{ \beta \} \vdash \lnot \alpha$.

And from $\Gamma \vdash \alpha$ we have also: $\Gamma \cup \{ \beta \} \vdash \alpha$.

This means that $\Gamma \cup \{ \beta \}$ is inconsistent, and thus $\Gamma \cup \{ \beta \} \vdash \gamma$, for every formula $\gamma$.

Title: Having problems this proof


问:
Suggest an Antique Math Book worth reading?
I'm not a math wizard, but I recently started reading through a few math books to prepare myself for some upcoming classes and I'm starting to really get into it. Then I noticed a few antique math books at a used bookstore and bought them thinking that, if nothing else, they would look cool on my bookshelf. But as it turns out, I enjoy both reading and collecting them. I find myself constantly browsing used book stores, thrift stores, antique stores ect. looking for the next book to add to my library.

So do you know of an antique book that you found interesting, helpful, or historically relevant?

(Just some insight- some of the books I have that I like are: The Laws of Thought by George Boole; Mathematical Methods of Statistics by Harald Cramer; and Introduction to Mathematical Analysis by F.L. Griffin. I've also enjoyed reading online about probablity, logic, and math history. But any area of mathematics is fine as I'm still discovering which areas interest me.)

答: What about Disquisitiones Arithmeticæ? There is at least one modern translation into English.


问:
generalization and the Poincaré and Brouwer theorem
Theorem:

There exists a continuous vector field without singularities on the sphere $S^{n}$ , if only if , $n$ is odd.

I'm really stuck on how to prove this.

答:
Neither are generally true.

If $\frac ab = -\frac 12$, then the right hand side is

$$-\left\lfloor-\frac ab\right\rfloor - 1 = -\left\lfloor\frac 12\right\rfloor - 1 = -0-1 = -1$$

For the first, as a counter example, consider $x = 0$. Then

$$\lfloor x \rfloor = 0 \ne -1$$

For the second, as a counter example, consider $x = -42$. Then

$$\lfloor x \rfloor = -42 \ne -1$$


问:
Specific parametrization of a sphere
I would like to parametrize the region
\begin{equation}
M=\{(x,y,z):x^2+y^2+z^2 \leq 1 \text{ with } z \leq x+y \}
\end{equation}
and calculate the flux $\iint_S F\cdot n \ dS$ where $S=\partial M$. The problem is'nt the flux but this parametrization. With spherical coordinates i know how to write the sphere, but the inequality
\begin{equation}
z\leq x+y
\end{equation}
stays
\begin{equation}
\cos(\varphi)\leq\cos(\theta)\sin(\varphi)+\sin(\theta)\sin(\varphi)
\end{equation}
and im stucked at this part. My geometric intuition tells me that $\theta$ is dependent of $\varphi$, but  how write this? Probably in the flux part we are going to use divergence theorem, because the surface is an sphere with lid $z=x+y$, so we need a parametrization here in order to write $\iiint_M div(F) \ dV$, right?

答:
Your parametrization is fine. The bounds would be

$$\int_0^{2\pi}\int_{\cot^{-1}\left(\sin\theta+\cos\theta\right)}^{\pi}F\cdot n d\varphi d\theta$$

Doing it the other way would split the integral in two.

Title: Specific parametrization of a sphere


问:
In how many ways can you choose cleaning managers when person A refuses to clean together with B
A group of 8 people meet every week and bake together. Three people are responsible for cleaning each time. In how many ways can you choose cleaning managers when person A refuses to clean together with B, and person C offers to clean on the condition that neither A nor B helps.

I have thought that since person A can't work wtih either A or C, they can work in 154 = 20
different groups but then I realized that I can apply this to both C and B because A, B and C will never work together.

But I'm not sure if it's correct because then there were only be 3 * 20 = 60 ways

答:
Break into cases:

$A$ works (And so $B$ doesn't and $C$ doesn't)

$B$ works (And so $A$ doesn't and $C$ doesn't)

$C$ works (and so $A$ doesn't and $B$ doesn't)

None of $A,B,$ nor $C$ work.

 $3\cdot \binom{5}{2} + \binom{5}{3} = 3\cdot 10 + 10 = 40$


问:
Why is this QCQP non-convex?
Why is the following QCQP non-convex?

$$\begin{array}{ll} \underset{x,y \in \Bbb R}{\text{minimize}} & (x+2)^2 + (y+2)^2\\ \text{subject to} & x y \ge 4\\ & x \le 4\\ & y \ge 2\end{array}$$

The objective function is convex and the set of feasible solutions is also convex (follows from the graph).

答:
The problem is not convex because it is not written as $\min \{ f(x) : g_i(x) \leq 0, i=1,2,\ldots,m \}$ where $f$ and $g_i$ are convex functions.

The feasible region is convex as you point out. You can obtain a convex formulation by applying a logarithmic transformation to the nonconvex constraint: $\log x + \log y \geq \log 4$. This constraint implicitly restricts $x$ and $y$ to positive reals, which looking at the other constraints does not reduce the feasible region.

Title: Why is this QCQP non-convex?


问:
Construct a basis from 2 vector sets
Is it possible to construct a basis for R^3 by choosing 2 vectors from one set and 1 vector from another set?

I know that the basis can be constructed if 3 vectors are linear independent but I don't know if you can take from different sets.

答:
A basis is defined a subset of a given vector space which is linearly independent and spans the entire vector space. You do not need to care about where the elements of the basis came from, as when you construct a basis you are always creating a new set of elements. For example, if we're considering $\mathbb{R}^3$, the standard basis is given by
$$
\alpha  =\{(1,0,0), (0,1,0),(0,0,1)\}
$$
You could think of this as taking two vectors from the set $\{(1,0,0),(1,1,1),(0,0,1)\}$ and one vector from the set $\{(2,2,3),(0,1,0)\}$ but it doesnt matter. The important part is that your basis vectors lie in the underlying vector space $\mathbb{R}^3$.

Of course this doesn't mean that given at least three vectors you can construct a basis, it just depends on what the vectors are.

Title: Construct a basis from 2 vector sets


问:
Is the set of all matrices $A$ that have $Tr(BAB)=0$ ($B$ is any fixed matrix) a vector space?
I'm really struggling to write the $tr(BAB)$ with the formal definition (sigma and products).
I would appreciate it if someone can help me with how I approach questions like these with traces and products of matrices.
Thanks

答:
To show that something is a vector space you need it to be closed under addition an scalar multiplication.  So assume that $\operatorname{tr}(BAB)=\operatorname{tr}(BA'B)=0$.  Let us use properties of traces (but not the definition):
$$\operatorname{tr}(B(A+A')B)=\operatorname{tr}(BAB+BA'B)=\operatorname{tr}(ABA)+\operatorname{tr}(BA'B)=0+0=0.$$

Similarly, if $c$ is a scalar, we have
$$\operatorname{tr}(B(cA)B)=\operatorname{tr}(c(BAB))=c\operatorname{tr}(BAB)=c\cdot 0 = 0.$$

The first calculation tells us that our space is closed under sums.  The second tells us that our space is closed under scalar multiplication.


问:
How to evaluate $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ without using antiderivative?
Someone gives a solution here:

\begin{align*} \int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}&=\int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty} (-\cos\alpha\cos\beta)^n{\rm d}\alpha\\ &=\sum_{n=0}^{\infty}(-\cos\beta)^n\int_0^{\frac{\pi}{2}} \cos^n\alpha{\rm d}\alpha\\ &=\sum_{n=0}^{\infty}(-\cos \beta) ^n\frac{\sqrt{\pi}\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{2\Gamma\left(\frac{n}{2}+1\right)}\\  &=\frac{\pi-2\arcsin\cos \beta}{2\sqrt{1-\cos^2\beta}}\\ &=\frac{\beta}{\sin \beta}.   \end{align*}

Is it correct? How to obtain the fourth equlity?

答:
$\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ =  $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+k\cos\alpha}$

then use $cos\alpha$ = ($cos$ ^2)$\alpha/2$ - $(sin$^2)$\alpha/2$.

divide numerator and denom by ($cos$ ^2)$\alpha/2$

then take tan$\alpha/2$ = t

then u can get a quadratic in denominator which can be factorized, get two fraction.


问:
Wave rider function
I'm looking for an example of a 'wave-like' function $f : \mathbb{R} \longrightarrow \mathbb{R}$ with the following properties:

deterministic

smooth (continously differentiable)

periodic but with varying period lengths $p_i$, $p_i \neq p_j$

mean period length $p=\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{i=-n}^n p_i$ is finite but not $0$ (periods don't get constantly larger or smaller)

I tried various combinations of sine functions but that seems more tricky than I thought.

A non-trivial example might be the Riemann zeta function (normalized to have unit average spacing), i.e., $f(x) = \text{Re}\zeta(\frac{1}{2}+\mathsf{i} \delta(x))$.

Background: The question is related to how random the roots of a deterministic function can be?

答:
I think $\displaystyle f(x) = \sin\big(x + 10\sin\big(x^{\frac{3}{5}}\big)\big)$ should work.

It has a vertical tangent at $x=0$ because $x^{\frac{3}{5}}$ is not differentiable there, but it satisfies all of your other conditions. In particular, the period varies but never goes to $0$. (I think it averages out to $2\pi$ because the second term's influence lessens over time, but I'm not sure.) You could fix the vertical tangent issue by using a differentiable-everywhere function in place of $x^{\frac{3}{5}}$, which would have to be sublinear so the period doesn't go to $0$.

On a sidenote, I found this function while playing around, which I think looks pretty cool.

$f(x) = \sin\big(x+\sin\big(x^{2}\big)\big)$

Title: Wave rider function


问:
Let $a >0$. Find the area bounded by $x+y=a$ and $\sqrt{x}+\sqrt{y} = \sqrt{a}$.

Let $a >0$. Find the area bounded by $x+y=a$ and $\sqrt{x}+\sqrt{y} = \sqrt{a}$.

Graphing this it's a line $x+y=a$ on top of the curve $\sqrt{x}+\sqrt{y} = \sqrt{a}$. So solving for $y$ in both equations I have that $y= a-x$ and $y= (\sqrt{a}-\sqrt{x})^2$

So the integral becomes $$\int _0^a\int _{\left(\sqrt{a}+\sqrt{x}\right)^2}^{a-x}1 \ dy \ dx = \int _0^a\left(-2\sqrt{a}\sqrt{x}-2x\right) \ dx = -\frac{4}{3}\sqrt{a}a^{\frac{3}{2}}-a^2$$

but this seems to have a negative area, what might be happening here?

答:
The first thing I did was graph the two equations.  Of course x+ y= a is the straight line through (a, 0) and (0, a).  $\sqrt{x}+\sqrt{y}= \sqrt{a}$ is more "interesting".  When x= 0 we have $\sqrt{y}= \sqrt{a}$ so $y= a$.  When y= 0 we have $\sqrt{x}= \sqrt{a}$ so $x= a$.  This graph also passes through (a, 0) and (0, a).  In the "middle", with x= y, $\sqrt{x}+ \sqrt{y}= \sqrt{x}+ \sqrt{x}= 2\sqrt{x}= \sqrt{a}$ so 4x= a.  x= a/4.  (a/4, a/4) is a point on the graph so the graph of $\sqrt{x}+ \sqrt{y}= \sqrt{a}$ lies below the graph of $x+ y= a$.  Of course, from $x+ y= a$, y= a- x and from $\sqrt{x}+ \sqrt{y}= \sqrt{a}$, $\sqrt{y}= \sqrt{a}- \sqrt{x}$ so $y= (\sqrt{a}- \sqrt{x})^2= a- 2\sqrt{ax}+ x$.

We want to integrate $y_1- y_2= a- x- (a- 2\sqrt{ax}+ x)= 2\sqrt{ax}- 2x$ from x= 0 to x= a:
$\int_0^a 2a^{1/2}x^{1/2}- 2x dx$


问:
Business sales math
Ok. I am trying to put together a 5-year revenue projection.  I only have the following data:

Units Sold in Year 1: 20

Units Sold in Year 2: 80

Units Sold in Year 3: 200

Now I need to use this data to figure out:

What monthly growth rate to use for each month from Year 1 to Year 5 to project monthly unit sales?

Similarly, growth rate to use for each quarter?

The 5-year compound annual growth rate?

答:
As Arturo pointed out, using uniform growth here is probably not what was intended. Assuming that the growth rate halves every year seems more plausible. Also, the question 1 seems to indicate that you should find a different growth rate for each month (and hence quarter).

My intuitive answer would be the following:

Since the growth rate seems to halve every twelve months, use the formula for half-life:

$$ g = g_{0} \left( \frac{1}{2} \right) ^{\frac{t}{12}}$$

where $g$ is your growth rate, and $g_{0}$ is your starting value of growth (in this case, 600) and $t$ is the time in months..

You can compute the compound rates from there.

Then again, this is just a gist. The question seems quite vague and leaves many room for different assumptions.

Title: Business sales math


问:
Showing set of tensored states span a space
I have the four states
$$
\lvert1\rangle \lvert1\rangle - \lvert0\rangle \lvert0\rangle \\
i\lvert1\rangle \lvert1\rangle + i\lvert0\rangle\lvert0\rangle \\
\lvert0\rangle \lvert1\rangle + \lvert1\rangle \lvert0\rangle \\
\lvert0\rangle \lvert1\rangle - \lvert1\rangle \lvert0\rangle
$$

What is a quick and easy way to show they are an orthogonal basis in $\mathbb{C^2}\otimes \mathbb{C^2}$. I've got the simple orthogonal bit down which I guess proves independence, but how can I show the span of the set?

答:
$$
\begin{eqnarray}
|1\rangle|1\rangle-|0\rangle|0\rangle \\
i|1\rangle|1\rangle+i|0\rangle|0\rangle \\
|0\rangle|1\rangle+|1\rangle|0\rangle \\
|0\rangle|1\rangle-|1\rangle|0\rangle \\
\end{eqnarray}
$$
forms an orthogonal basis, since the (scaled) matrix
$$
U=\frac{1}{\sqrt{2}}\pmatrix{
-1 &0&0&1\\
i&0&0&i\\
0&1&1&0\\
0&1&-1&0
}
$$
which transforms the standard basis to your (entangled) basis, is unitary (check $U^\dagger U\overset{!}{=}\bf 1$) and this preserves orthogonality.</p>


问:
Integrating in cylindrical coordinates and finding surface area
I have a couple study guide involving finding integrals using cylindrical coordinates and finding one using a scalar surface. The main thing I struggle with is finding bounds.

One question asks me to express a triple integral in cylindrical coordinates for the volume of the sphere $x^2+y^2+z^2=16$ contained within the cylinder $(x-2)^2+y^2=4$.

The other questions asks me to find the surface area of the paraboloid $z=5-2x^2-2y^2$ that lies above the plane z=-13. The main thing I have trouble with is the bounds. Can someone explain how to do these problems for me?

I know for the first one that if the cylinder is expressed that way I can expand $(x-2)^2$ to $x^2-4x+4$ and then convert the sphere integrand and integral bounds to cylindrical. I was just asking for someone to help out because the professor did not give answers to the study guide.

答:
For the first one, your sphere is centered at the origin with radius of $4$ whereas your cylinder is along $z$ axis, centered at $(2, 0, 0)$ and radius of $2$.

Using cylindrical coordinates,

Using $x = r \cos \theta, y = r \sin \theta \, , \, $your cylinder $x^2 - 4x + y^2 = 0 \,$ can be written as,

$r = 4 \cos \theta \, ( - \pi/2 \leq \theta \leq \pi/2)$. Please note the bounds of $\theta$.

Bounds of $z$ are spherical caps, $z = \pm \sqrt{16 - x^2 - y^2} = \pm \sqrt{16 - r^2}$

So bounds for your integral are

$-\sqrt{16-r^2} \leq z \leq \sqrt{16-r^2}$

$0 \leq r \leq 4\cos \theta$

$-\pi/2 \leq \theta \leq \pi/2$

For the second one,

Your parametrized surface is $z = f(x,y) = 5 - 2x^2 - 2y^2 \,$ which is an inverted paraboloid with vertex at $(5, 0, 0)$. As we need to find surface area above plane $z = -13$, we have

$-13 \leq z \leq 5, 0 \leq x^2 + y^2 \leq 9$. So the projection in $XY$ plane is a disc $0 \leq r \leq 3$.

The surface area is given by $\iint_D \sqrt{1+f_x^2 + f_y^2} \, dA \,$ where $dA$ is the projection of the surface in the $XY$ plane.

Taking derivative wrt $x, y, \, f_x = -4x, f_y = -4y$

In polar coordinates, $dS = \sqrt{1+f_x^2 + f_y^2} \, dA = \sqrt{1 + 16r^2} \, r \, dr \, d\theta$

Can you take it from here?


问:
Taking derivative on both sides of an equation
What theorem or rule is used for this:

$x = f(\theta)$

$dx = f'(\theta) d\theta$

I would have expected $dx = f'(\theta)$

Thanks

答:
x is a function of theta.

So a tiny change in x (dx) must equal the instantaneous rate of change (slope) of the function of theta times the tiny change of theta (dtheta).

You can think of it as literally just the slope formula with infinitesimals.


问:
Why is boundary information so significant? -- Stokes's theorem
Why is it that there are so many instances in analysis, both real and complex, in which the values of a function on the interior of some domain are completely determined by the values which it takes on the boundary?

I know that this has something to do with the general version of Stokes's theorem, but I'm not advanced enough to understand this yet -- does anyone have a (semi) intuitive explanation for this kind of phenomenon?

答:
If you want to compute rate at which something is growing inside a surface (say, ppm of some molecule in a fluid, or electrons, etc) then it is useful to know the rate at which those particles are flowing through the boundary of that surface.

Consider, for example, electric current. The current to and from the boundary of a surface indicates how much the charge inside the surface is building or diminishing.


答:
To complement Aaron's nice answer, one could also notice that the map

$$f \colon \mathrm{Mat}(n \times n, k) \to k, \ \ \ \ \ A \mapsto \mathrm{tr}(BAB)$$

is linear. (Here, $k$ is any field and $\mathrm{Mat}(n \times n, k)$ is the space of $n \times n$-matrices with entries in $k$). Of course, one proves this with similar arguments as in Aaron's answer.

Then your desired space is just the kernel of $f$, hence a vector subspace of $\mathrm{Mat}(n \times n, k)$.

Title: Is the set of all matrices $A$ that have $Tr(BAB)=0$ ($B$ is any fixed matrix) a vector space?

Body:
I'm really struggling to write the $tr(BAB)$ with the formal definition (sigma and products).
I would appreciate it if someone can help me with how I approach questions like these with traces and products of matrices.
Thanks


问:
Prove $ \forall x\in\mathbb{R} : |\sin{x}|<|x|$ geometrically
For $x\in(-\frac{\pi}{2}, -\frac{\pi}{2})\setminus \{0\}$ we can say that since the area of sector of circle is greater than the triangle (ABC) area I have $2|\sin{x}|<2|x|$. But what's about x in the 2nd and 3rd quadrant?
Look at the figure to easily understand what I am saying (Note: I have drawn a circle with a radius $>1$ to have a larger circle, but I am referring to a unit circle)

答:
Area of the triangle is $$S_\triangle = \frac{1\cdot 1\cdot \sin x}{2}$$
Area of the sector is $$S_\text{sec} = \frac{1^2\cdot x}{2}$$
Since it is vivid from the picture below that $S_\text{sec} > S_\triangle$, we have
$$\sin x < x$$
As has already been noted in the comment, you just need to verify only for $x \in [-1, 1] \subset (-\pi/2, \pi/2)$. The above argument can easily be extended to $\forall x \in [-1, 1]$ (note: equality occurs when $x=0$).

$\quad\quad\quad\quad\quad\quad\quad~$


问:
Prove that the transformation T over a unit vector is also a unit vector
Let the $T : \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation. How can I prove that the transformation T over a unit vector is also a unit vector? Does a linear transformation always preserve the distance?

I know that there must exist a matrix $A$ such that $T(u) = A u$ and $||u||^2 = u \cdot u =1$.

$$||T(u)||^2 = T(u) \cdot T(u) = Au \cdot Au$$

Not sure how to finish this.

答: It’s not true in general. Counter example: $Tx=2x$.


答:
Linear transformations are affine, which means they preserve colinearity, parallelism, and origin, but they need not preserve distance. For example the matrix given by
\begin{bmatrix}
k & 0\\
0 & 1
\end{bmatrix}
Corresponds to the map $T:\mathbb{R}^2\to\mathbb{R}^2$ which stretches everything by $k$ along the $x$ axis. You can verify that this does not preserve unit vectors by considering the unit vector along the $x$ axis (unless $k = \pm 1$).

Title: Prove that the transformation T over a unit vector is also a unit vector

Body:
Let the $T : \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation. How can I prove that the transformation T over a unit vector is also a unit vector? Does a linear transformation always preserve the distance?

I know that there must exist a matrix $A$ such that $T(u) = A u$ and $||u||^2 = u \cdot u =1$.

$$||T(u)||^2 = T(u) \cdot T(u) = Au \cdot Au$$

Not sure how to finish this.


问:
Multiple choice question Real analysis

Answer is given as option d.
As per the definition of the function the function is infinitely differentiable on $\mathbb{R}$.But I want to know why the function is not analytic.How we know it is not analytic?

thanks in advance!!

答: If you compute all derivatives at $x=0$, you will notice that they are all $=0$. An anylytic function with these derivatives would equal the corresponding power series in a neighbourhood of $0$, i.e., would be constantly $=0$ in that neighbourhood.

Title: Multiple choice question Real analysis


问:
How to show that if $a^2+b^2+c^2=1$ then $\frac{-1}{2}\le ab+ac+bc \le 1$
How to show that if $a^2+b^2+c^2=1$ then:

$$\frac{-1}{2}\le ab+ac+bc \le 1$$

From the assumption I see that:

$$-\frac{1}{2}\le\frac{\left(a+b+c\right)^{2}-1}{2}=\frac{\left(a+b+c\right)^{2}-\left(a^{2}+b^{2}+c^{2}\right)}{2}=ab+ac+bc $$
On the other hand $a^2=1-(b^2+c^2)<1$,then same does hold for $b^2$ and $c^2$ which shows that $-1<a<1$ and so :$$-3<a+b+c<3$$ $$\implies ab+ac+bc =\frac{\left(a+b+c\right)^{2}-1}{2}<4$$

But it's does not what I've been asked to prove,so how can I show that?

答:
Hint:
For the 2 inequalities you want to prove, expand the left hand sides of:
\begin{align}
(a+b+c)^2&\geq 0,\\
(a-b)^2+(b-c)^2+(c-a)^2&\geq 0.
\end{align}
And remember to use $a^2+b^2+c^2=1$.


问:
Convergence of a sequence s.t. $u_{n+1}-u_n\leq \frac{1}{(n+1)^2}$
I'm looking for an simple proof of the following fact :

Let $(u_n)_{n\in \mathbb N}$ be a sequence of non-negative real
numbers satisfaying :  $$\forall n\in \mathbb N, \quad u_{n+1}-u_n\leq \dfrac{1}{(n+1)^2}$$ Then $(u_n)$ is convergent.

I know that $(u_n)$ is bounded since $$\forall n\in \mathbb N, u_n\geq 0 \quad \text{ and } \quad u_n=\sum_{k=0}^{n-1}\left(u_{k+1}-u_k\right) + u_0 \leq \sum_{k=1}^{n}\dfrac{1}{k^2} + u_0\leq \dfrac{\pi^2}{6}+u_0.$$

One could consider the set $A=\{n\in \mathbb N \ | \ u_{n+1}-u_n\geq 0\}$ in order to argue about increasing or decreasing subsequences, but it seems to technical.

Any ideas ?

答: $$u_{n+1}-u_n\leq \dfrac{1}{(n+1)^2}<\frac1{n(n+1)}=\frac1n-\frac1{n+1}$$ implies $$u_{n+1}+\frac1{n+1}-\left(u_n+\frac1n\right)<0,$$ i.e. $u_n+\frac1n$ is a decreasing positive sequence, and thus convergent.


问:
Finding a contradiction for inequalities
Assuming that $a_1 \leq b_1\leq c_1 \leq d_1 \le n_1$ and $a_2 \leq b_2\leq c_2 \leq d_2 \le n_2$ and $(a_1 + a_2) < (b_1 + b_2) < (c_1 + c_2) < (d_1+d_2)$ (and all the numbers are natural and positive) can we find a contradiction in the following four inequalities:

$n_2 a_1 \ge a_2 (n_1-1)$,$n_2 b_1 \ge b_2 (n_1-1)$, $ (n_2-1) c_1 \le c_2 (n_1)$ and $d_1 n_2 \ge d_2(n_1-1)$?

If yes, could you please give a hint about how to proceed?

Body:
No.

Works e.g. with $a_1=2$, $a_2=1$, $b_1=3$, $b_2=1$, $c_1=3$, $c_2=10000$.


问:
Is any ring map between two $K$- extensions necessarily $K$-linear?
This may sound really obvious but I was considering the following:

Suppose $K$ is a field and $M,L$ are two extensions of $K$ (often denoted by $M|K, L|K$). If I consider a ring homomorphism from $M$ to $L$, say $\phi:M\to L$. Is it necessarily true that $\phi$ is $K$-linear? That is, is it necessarily true that $\phi(km)=k\phi(m) \forall k\in K,m\in M$?

答:
It is not. Take, for instance, $K=\mathbb Q(\sqrt2)$ and (as a trivial example) $M=L=K$. Now there is a ring homomorphism $M\longrightarrow L$ with $a+b\sqrt2\mapsto a-b\sqrt2$. This homomorphism doesn't satisfy $\phi(k)=k$ for all $k\in K$, so it's not $K$-linear.

But even if you want less trivial examples, take $K$ as above and $M=\mathbb Q(\sqrt2,\sqrt3),~L=\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$. Then we have a homomorphism $\phi:M\longrightarrow L$ given by
$$a+b\sqrt2+c\sqrt3+d\sqrt6\mapsto a-b\sqrt2+c\sqrt3-d\sqrt6.$$
This is also not $K$-linear.

The fact that $K$-linearity isn't guaranteed is, by the way, the reason why we explicitly define the Galois group of a Galois extension as $\operatorname{Gal}(M\vert K):=\operatorname{Aut}_K(L)$, that is, the automorphisms $L\longrightarrow L$ which fix $K$ elementwise. In other words, the $K$-linear automorphisms of $L$.


问:
Non-linear transformation of independent standard normal distributions
I am attempting to find the p.d.f of the following random variable $V$:

Given $X_1$ and $X_2$ follow standard normal distributions and are independent. I want to find the p.d.f of $V = X_1^2 + X_2^2$.

My idea is to do let $Z = X_1^2$ and $Y=X_2^2$ and then calculate the p.d.f of $V = Z+Y$.

Proposed Method:

$F_Z(z) = P(Z \le z) = P(X_1^2 \le z) = P(-\sqrt{z} \le X_1 \le \sqrt{z}) = 2\Phi(\sqrt{z}) -1$

We also have a similar relationship with $Y$.

Then we get that $F_V(v) = P(V \le v) = P(Z+Y \le v) = P(Z \le v-y)$ - This is the step I am not sure of. If I can do this then I get that $F_V(v) = 2\Phi(v-y) -1 \implies f_V(v) = 2\phi(v-x_2^2) \implies f_V(v) = 2\phi(x_1^2)$

This seems that it most likely is wrong. Where am I going wrong and is there a way to do this using moment generating functions?

Edit: I am incorrect as when differentiating I have not applied the chain rule, and I have not square rooted the contents of the phi

答:
$V=X_1+X_2$ is the definition of a known rv: a chi-square rv with 2 d.0.f

To prove this is very easy:

First, with a simple transformation verify that $X_1^2\sim \chi_{(1)}^2$

Second, using the properties of Gamma distribution prove which is the distribution of the sum (or via MGF, Characteristic function)

To prove that $X_1^2\sim \chi_{(1)}^2$ you can do as follows

Set $X\sim N(0;1)$ and let's derive the distribution of

$$Y=X^2$$

$$F_Y(y)=\mathbb{P}[Y\leq y]=\mathbb{P}[-\sqrt{y }\leq X\leq \sqrt{y}]=\Phi[\sqrt{y}]-\Phi[-\sqrt{y}]$$

Derivating you get

$$f_Y(y)=\phi(\sqrt{y})\frac{1}{2\sqrt{y}}+\phi(-\sqrt{y})\frac{1}{2\sqrt{y}}=\frac{1}{\sqrt{y}}\phi(\sqrt{y})$$

That is

$$f_Y(y)=\frac{1}{\sqrt{y}\sqrt{2\pi}}e^{-\frac{y}{2}}=\frac{\Big(\frac{1}{2}\Big)^{\frac{1}{2}}}{\Gamma\Big(\frac{1}{2}\Big)}y^{\frac{1}{2}-1}e^{-\frac{y}{2}}$$

Thus

$$Y\sim Gamma\Bigg(\frac{1}{2};\frac{1}{2}\Bigg)=\chi_{(1)}^2$$


问:
How are Jacobians of genus $3$ curves different from one another?
There are two types of smooth projective (complex) curves of genus $3$: plane quartics, and hyperelliptic curves. The Torelli morphism $M_3\to A_3$, assigning a curve to its (principally polarized) Jacobian, is bijective on points. Thus $A_3$ consists of Jacobians of genus $3$ curves. Because of the "partition" of $M_3$ as {plane quartics} $\amalg$ {hyperelliptics}, I would expect this to be reflected somehow in $A_3$. Whence my questions:

Question 1. What are the main differences between Jacobians of plane quartics and of hyperelliptic curves?

(I am intentionally vague here, as I am curious about any sort of difference).

It would be also interesting to know if the deformation theory of the curve inside its Jacobian is sensibly different according to the curve itself; this leads to a second question:

Question 2. (Deformations inside the Jacobian) What do the normal bundles $\mathcal N_{C/Jac(C)}$ look like, when $C$ is a plane quartic or a hyperelliptic curve, respectively?

Thanks for any clue on this.

答:
For Question 2. From the conormal bundle sequence
$$0\to\mathcal N^{\vee}\to \Omega_{Jac(C)}\to \omega_C\to 0$$
we see that we can identify $\mathcal N^{\vee}$ with the kernel/syzygy bundle $M_{\omega_C}$ given by the exact sequence
$$0\to M_{\omega_C}\to H^0(C,\omega_C)\to\omega_C.$$
This bundle is semistable of slope $-2$ and if $g\geq 3$ it fails to be stable precisely for hyperelliptic curves. See (3.4)Corollary in https://www.imsc.res.in/~kapil/papers/chap1/index.html

To see the instability appear for hyperelliptic curves, it is enough to observe sections of $M_{\omega_C}\otimes{\mathcal O}_C(H)$, where $H$ is the hyperelliptic pencil giving the $2:1$ map to $\mathbb P^1$, such that $h^0(C,H)=2$. Twisting in the sequence above and taking sections, we have
$$0\to H^0(C,M_{\omega_C}\otimes{\cal O}_C(H))\to\mathbb C^{2g}\to\mathbb C^{g+1}.$$


问:
If $f(x)$ is equal to $x$ or $1-x$ when $x$ is rational/irrational, then $f$ is continuous only at $x=1/2$
How can I resolve the task below:

Given $f(x)=
\begin{cases}
 x, &x\in \mathbb{Q}\text{  }\\
 1-x, &x\notin \mathbb{Q}\text{ (irrational)}
\end{cases}$, $0 \leq x \leq 1$.

Show $f(x)$ is continuous only in $x=\dfrac{1}{2}$.

答:
ncmathstudent's answer is good, but if you want to see an alternative proof:

Note that both $\mathbb Q$ and $\mathbb R\setminus \mathbb Q$ are dense in $\mathbb R$. Thus for any $r\in\mathbb R$ there is a unique continuous function on $\mathbb Q\cup \{r\}$ which agrees with $x$ on $\mathbb Q$ (which clearly must be $x$), and a unique continuous function on $\mathbb R\setminus \mathbb Q\cup \{r\}$ which agrees with $1-x$ on $\mathbb Q\setminus \mathbb R$ (which clearly must be $1-x$). Thus if $f(x)$ is continuous at $r$, we must have $x=1-x$ at $r$ so $r=1/2$. It remains to verify that $f(x)$ is continuous at $1/2$, which is easy.


问:
Exact solutions of $x$ in $\cos x(\pi -1)=\cos x\pi$?
I want to find an exact solution for $x>0$ which solves the following equation

$$\cos x(\pi -1)=\cos   x\pi$$

Mathematica gives an answer $x=\frac{4 \pi  c_1}{2 \pi -1},c_1\in\mathbb{N}$, but when I substitute this value, say for $c_1=1$, then $\cos x(\pi -1)-\cos   x\pi$ is very close to zero, but not exactly zero.

Then, can I use $x=\frac{4 \pi  c_1}{2 \pi -1},c_1\in\mathbb{N}$ as an exact solution of the given equation?

答:
Use
$$\cos\alpha-\cos\beta = 2\cdot\sin\frac{\alpha+\beta}{2} \cdot\sin\frac{\beta-\alpha}2$$


问:
Looking for an example of sequence of analytic functions with weird behaviour
Denote by $\mathbb{D}$ the open unit disk of the complex plane $\mathbb{C}$ and let $D = \{z_n\}_{n\geq 1}$ be a dense set on $\mathbb{D}$. I'm looking for a sequence of analytic functions $(f_n)$ on $\mathbb{C}$ such that converges uniformly to an analytic function $f$ on $\mathbb{C}\smallsetminus\overline{\mathbb{D}}$, converges pointwise on the dense set $D$ but does not converge on $\mathbb{D}\smallsetminus D.$ Can we construct such an example?

Thank you very much!

答:
No, if $f_n$ converges uniformly on $|z|=2$ then it does so on $|z|\le 2$.

$$\sup_{|z|\le 2} |f_n(z)-f_m(z)|\le \sup_{|z|=2} |f_n(z)-f_m(z)|$$


问:
How to prove $\mathbb{Z}[i]$ is a ring but not a field?
Let i be the imaginary unit of $\Bbb{C}$ and $\mathbb{Z}[i]:= \{a + ib, ab \in\mathbb{Z} \} $

So the task is to prove that this a ring regarding multiplication and addition in $\mathbb{C}$ which is commutative, free of zero divisors and posseses a 1 element.
I already found out that it is a ring and that the 1 Element is (1,0) and 0 Element is (0,0), so in both cases $i= 0$. So I guess this is like proving a trivial Ring.
I am just confused because I am not sure about the $i$. How would I complete the proof? I am stuck because I don't know how to prove that it is free of zero divisors and not a field.
Thank you for the help.

答:
Though there are simple direct proofs, it is worth emphasis that this is an immediate consequence of a fundamental property of integral extensions, i.e. extensions obtained by adjoining algebraic integers (roots of polynomials that are monic, i.e. lead coef $= 1).\,$ Namely, integral extension don't alter unit properties of the base ring, so a nonunit (i.e. noninvertible) element remains nonunit in the extension. In particular if the base ring is not a field then ditto for the integral extension. Let's prove this very simply for your special case.

Write $\,\Bbb I := \Bbb Z[i]\cong \Bbb Z[x]/(x^2+1)\,$ for the  Gaussian integers. Note every $\,w\in\Bbb I\,$ is integral over $\,\Bbb Z\,$ being a root of a monic polynomial $\,f(x)=(x\!-\!w)(x\!-\!\bar w) = x^2+bx+c  \in\Bbb Z[x].\,$ If an integer $n\in\Bbb Z$ is invertible in $\Bbb I\,$ then $\,w = n^{-1}\,$ is a root of such a quadratic so $\,n^{-2}+b\,n^{-1}+c = 0\,$ $\overset{\times\ n}\Rightarrow\, n^{-1} + b + c\,n = 0,\,$ so $\,n^{-1} = -b - c\,n \in \Bbb Z,\,$ i.e. $\,n\,$ was already invertible in $\,\Bbb Z.\,$ Thus nonunits in $\,\Bbb Z\,$ remain nonunits in $\,\Bbb I,\,$ e.g. $\,2\,$ is not invertible in $\,\Bbb I,\,$ so $\,\Bbb I\,$ is not a field.

Exactly the same simple proof generalizes to any integral extension, e.g. see here. Generally this method easily shows that

$\qquad$ If $\, E < D\,$ is an integral extension of domains then $\,E\,$ is a field $\!\iff\! D\,$ is a field.

This preservation of units in integral extensions mean that they faithfully preserve divisibility properties of the base ring, so e.g. we can deduce divisibility properties of integers from divisibility properties of Gaussian integers and other rings of algebraic integers (where proofs my be simpler because we may be able to reduce nonlinear problems to linear problems via factorizations, e.g. classical proofs of Fermat's Last theorem for small exponents). This would not be true if some nonunit changed to a unit in the extension, e.g. if $2$ became a unit in the extension then proofs by parity would break down in the extension, so it would be weaker from a divisibility standpoint.


问:
Prove an inequality in proof of Poincaré recurrence theorem.
I read this lecture made by Terence Tao.

(https://terrytao.wordpress.com/2008/01/30/254a-lecture-8-the-mean-ergodic-theorem/)

I tried to understand the proof of Poincaré recurrence theorem.

Everything is clear for me except one inequality:

Prove that $$\sum_{n=1}^{N}\sum_{m=1}^{N}\mu(E\cap T^{m-n}E) \le (\limsup_{n\to \infty}\mu(E\cap T^n(E))+o(1))N^2  $$

Here $\mu(E\cap T^{m-n}E)$ was originally replaced by $\mu(T^nE\cap T^mE)$ but it is written that this is the same and i agree with it.

I can also see that left-hand side is a sum of the measures $\mu(E\cap T^k(E))$, where $k=0,1,..N-1$

But I am able only to prove a different statement:
$$\sum_{n=1}^{N}\sum_{m=1}^{N}\mu(E\cap T^{m-n}E) \le (\sup_{n\le N}\mu(E\cap T^nE))N^2$$

According to this lecture, I have also one question:

How  theorem 1 in this lecture is connected with the original Poincaré recurrence theorem?(the theorem where one can find at Wikipedia page or somewhere else).

Regards

答:
First as you said one can simplifye the sums as follow $\sum_{m=1}^N \sum_{n=1}^N \mu(E \cap T^{m-n} E)= \sum_{k=N-1}^{N-1} (N -|k|) \mu(E \cap T^k E)$

Then because of the invariance of the measure $\mu(E \cap T^k E) = \mu(E \cap T^{-k} E)$

So $\sum_{k=N-1}^{N-1} (N -|k|) \mu(E \cap T^k E) = 2 \sum_{k=1}^{N-1} (N -k) \mu(E \cap T^k E) + N \mu(E)$

Our first upper bound : $ 2 \sum_{k=1}^{N-1} (N -|k|) \mu(E \cap T^k E) \leq 2 N \sum_{k=1}^{N-1} \mu(E \cap T^k E)$

Now we will justify the following lemma :

Let $x_n$ be a sequence of real beetween $0$ and $1$, we will note $x_+ = \lim \sup x_n$ then $\sum_{k=1}^{N} x_n \leq N x_+ +o(N)$

We have $\sum_{k=1}^N x_k = \sum_{k=1}^{\lfloor \sqrt{N} \rfloor} x_k + \sum_{k=\lfloor \sqrt{N} \rfloor+1}^N x_k \leq \lfloor \sqrt{N} \rfloor \max_{1 \leq k \leq \lfloor \sqrt{N} \rfloor} x_k + (N-\lfloor \sqrt{N} \rfloor -1)  \max_{ \lfloor \sqrt{N} +1 \rfloor \leq k \leq N} x_k$

Of course $\max_{ \lfloor \sqrt{N} +1 \rfloor \leq k \leq N} x_k = x_+ + o(1)$,
and because of the hypothesis $\max_{1 \leq k \leq \lfloor \sqrt{N} \rfloor} x_k \leq 1$

so $\sum_{k=1}^N x_k \leq \lfloor \sqrt{N} \rfloor + (N-\lfloor \sqrt{N} \rfloor -1)(x_+ +o(1)) = N x_+ + o(N) $

Applying this lemma we get $\sum_{k=1}^{N-1} \mu(E \cap T^k E) \leq N(\lim \sup \mu(E \cap T^k E) +o(1))$

and $\sum_{m=1}^N \sum_{n=1}^N \mu(E \cap T^{m-n} E) \leq 2 N^2 (\lim \sup \mu(E \cap T^k E) +o(1)) + N \mu(E)$

as $N \mu(E)=o(N^2)$ we have nearly our result.

I really don't know why I got a factor $2$..., if someone can check my proof.

To answer your second question

How theorem 1 in this lecture is connected with the original Poincaré recurrence theorem?(the theorem where one can find at Wikipedia page or somewhere else).

Tao use this equation to show that $\lim \sup \mu(E \cap T^n E) \geq \mu(E)^2$ that is given any set, after a long time, you are sure that the image of $E$ will eventually intersept $E$ enough to have measure $\mu(E)^2$.


问:
Solving an integral which results in elliptic integrals without never having heard of elliptic integrals
Last week I asked this question here and got a very well explained answer. Yet again I might say I am very very rusty on my mathematics, but in this case I've never even heard of elliptic integrals up to the point I saw this paper. In the paper they define:

$$\rho = r_m(\varphi) = \frac{1}{2}\lim_{n\to+\infty}\left(|\cos{\frac{m\varphi}{4}|^n}+|\sin{\frac{m\varphi}{4}}|^n\right)^{-\frac{1}{n}}=\frac{1}{2}\min{\left(\left|\sec{\frac{m\varphi}{4}}\right|,\left|\csc{\frac{m\varphi}{4}}\right|\right)}\label{1}(1)$$

Then they define that the length of the closed curve $C_m$ is given by
$$l_m =  \int_{0}^{2\pi} \sqrt{r_m^2(\varphi)+\left[\frac{dr_m(\varphi)}{d\varphi}\right]^2} \,d\varphi ~(3)$$

The authors say that by using (1) in (3) we can get the following:

$$l_m = \sqrt{8+\frac{m^2}{2}}+4\left[F\left(\frac{\pi}{4}\bigg|1-\frac{m^2}{16}\right)-E\left(\frac{\pi}{4}\bigg|1-\frac{m^2}{16}\right)\right] ~(4)$$

with $F\left(\cdot\bigg|\cdot\right)$ and $E\left(\cdot\bigg|\cdot\right)$ being the elliptic integrals of first and second kind, respectively. I've been able to use this formulation already to get some of the results I want, but I want to understand how the authors got there as it is quite puzzling to me.

Thank you in advance

答:
The curve is composed of $2m$ copies of the patch from $0$ to $\frac\pi m$, over which $r_m(\varphi)=\frac12\sec\frac{m\varphi}4$. Thus, expanding the derivative, we have
$$l_m=2m\int_0^{\pi/m}\sqrt{\frac14\sec^2\frac{m\varphi}4+\frac{m^2}{16}\tan^2\frac{m\varphi}4\sec^2\frac{m\varphi}4}\,d\varphi$$
Substitute $u=\frac{m\varphi}4$ and simplify:
$$=\int_0^{\pi/4}\sqrt{(\sec^2u)(16+m^2\tan^2u)}\,du=4\int_0^{\pi/4}\frac1{\cos^2u}\sqrt{1-\left(1-\frac{m^2}{16}\right)\sin^2u}\,du$$
Define $1-\frac{m^2}{16}=M$ for brevity. We can rewrite the integrand as
$$\frac{\sec^2u-M\tan^2u}{\sqrt{1-M\sin^2u}}=\frac{M-M\sec^2u+\sec^2u}{\sqrt{1-M\sin^2u}}=\frac{1-(1-M\sin^2u)+M\cos^2u-(M-1)\sec^2u}{\sqrt{1-M\sin^2u}}$$
$$=\frac1{\sqrt{1-M\sin^2u}}-\sqrt{1-M\sin^2u}+\frac{M\cos^2u-(M-1)\sec^2u}{\sqrt{1-M\sin^2u}}$$
The first two parts we recognise as the integrands of the $F$ and $E$ elliptic integrals respectively; their parameter is $M$ and the first argument is the upper limit of the last integral we had before we embarked on splitting the integrand, i.e. $\frac\pi4$. The integral of the last part is $(\tan u)\sqrt{1-M\sin^2u}$.

Finally we have
$$l_m=4F(\pi/4|M)-4E(\pi/4|M)+4[(\tan u)\sqrt{1-M\sin^2u}]_0^{\pi/4}=4F(\pi/4|M)-4E(\pi/4|M)+\sqrt{8+\frac{m^2}2}$$


答: Hints: Arclength (with units of distance) is the product of the radius $1$ and the angle $x$. Show how this arclength is longer than the altitude by showing how on a Euclidean plane, a line segment is the shortest path between two points.

Title: Prove $ \forall x\in\mathbb{R} : |\sin{x}|<|x|$ geometrically

Body:
For $x\in(-\frac{\pi}{2}, -\frac{\pi}{2})\setminus \{0\}$ we can say that since the area of sector of circle is greater than the triangle (ABC) area I have $2|\sin{x}|<2|x|$. But what's about x in the 2nd and 3rd quadrant?
Look at the figure to easily understand what I am saying (Note: I have drawn a circle with a radius $>1$ to have a larger circle, but I am referring to a unit circle)


问:
Solving coupled system of ODEs with python
I have to numerically solve a coupled system of ODEs of the following form:
$$\begin{align}
\begin{cases}
\dot{c}(t) = R(t)f(t)\\
\dot{R}(t) = R(t)G(t),
\end{cases}
\end{align}$$
where $c(t), f(t) \in \mathbb{R}^3, R(t), G(t) \in \mathbb{R}^{3 \times 3}$.

I usually solve ODEs with solve_ivp from scipy.integrate and matrix valued ODEs with the wrapper from the odeintw package. Is there a wrapper for coupled systems too?

答:
Are you thinking about some general wrapper or of how to construct your own wrapper like

```<code>def odesystem(t,u):
    c, R = u[:3], u[3:12].reshape([3,3])
    dc = R.dot(f(t))
    dR = G(t).dot(R)
    return np.concatenate([dc, dR.flatten()])
</code>```

You could also use odeintw with a $4\times 3$ array as state object, where then c is the first row and R the second to fourth row, c,R=u[0],u[1:4].

Body:
Are you thinking about some general wrapper or of how to construct your own wrapper like

def odesystem(t,u):
    c, R = u[:3], u[3:12].reshape([3,3])
    dc = R.dot(f(t))
    dR = G(t).dot(R)
    return np.concatenate([dc, dR.flatten()])

You could also use odeintw with a $4\times 3$ array as state object, where then c is the first row and R the second to fourth row, c,R=u[0],u[1:4].


问:
What is the chance that an infinite series of random numbers will, at some point, repeat all previously generated numbers in that order?
The main reason of this post is to provide a writeup of what I think is a correct solution to this problem from the r/math subreddit. I'll quote the poster to phrase the problem:

What is the chance that an infinite series of random numbers will, at some point, repeat all previously generated numbers in that order?

While there was some confusion in the comments, I believe what the poster means is something like the sequences below:
$$7,7,...$$
$$3,2,3,2,...$$
$$7,8,6,7,8,6,...$$
Where the dots afterward are arbitrary as the condition has already been fulfilled. If we start writing out a sequence of digits in this way, where the probability of selecting any digit is equal, what is the probability that something like this will happen?

I'll post my solution below. Feel free to comment or correct me.

答:
In base 2, the answer is

$$
\sum_{i=1}^{\infty}\frac{a(i)}{4^i}
$$

where $a(i)$ is OEIS A216958. Per comments, there is no known closed solution or recurrence for that sequence, so it seems unlikely there will exist one for this sum.

The 20th partial sum of the above is ~ 0.72995609.


问:
(Multiplicative) inverse of $\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in\mathbb Q[\sqrt[3]{7}]$
Set $\mathbb Q[\sqrt[3]{7}] = \{F(\sqrt[3]{7}) \mid F ∈ Q[x]\}$ is a field (with the usual addition and the usual multiplication).

Calculate the (multiplicative) inverse of
$$\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in \mathbb Q[\sqrt[3]{7}]$$

Note: Application of the Euclidean algorithm to the polynomials $x^2 + 3x + 1$ and $x^3 - 7$ could help.

Attempt:

I know there is a multiplcative inverse $\beta$ with $\alpha\beta=1$, and that one should exist in $\mathbb Q[\sqrt[3]{7}]$, but do not know how I would go about expressing it in any form simpler than $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$.

How can I determine a simpler way to express the value of $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$?

答:
You actually have two possibilities:

Either, denoting $x=\sqrt[3]7$, you try to find a linearcombination $ax^2+bx+c$ such that
$\;(ax^2+bx+)(x^2+3x+1)=1$, which leads to solving the linear system
$$\begin{cases}a+3b+c=0\\7a+b+3c=0\\21a+7b+c=1\end{cases},$$
which can be solved  finding the reduced row echelon form of the augmented matrix
$$\left[\begin{array}{rrr|l}1&3&1&0\\ 7&1&3&0\\ 21&7&1&1\end{array}\right].$$

Or, extending a bit the hint, you apply the extended Euclidean algorithm to the polynomials $X^3-7$ and $X^2+3X+1$, which are coprime, to obtain a Bézout's relation
$$u(X)(X^2+3X+1)+v(X)(X^3-1)=1,$$
which, substituting $x$ to $X$, shows the inverse of $x^2+3x+1$ is $u(x)$.


问:
Probability of an ant going from one point to another on a lattice.
I found this problem on brilliant.org. I get this kind of doubt a lot, in similar problems.
(This is my 1st question here, please go easy on me)

An ant stands on point A of the lattice below.

This is the lattice image

Every minute, the ant selects an adjacent point uniformly at random and moves to it. For example, if the ant is on point B, then it moves to the points A, C, or G with probability 1/3 each.

What is the probability that the insect moves to point D in exactly 2 moves from the point shown?

If I find the product of probabilities (A->G->D), the answer I get is 1/3*1/6 = 1/18

But if I try to count the events, I get the probability to go from A to D in 2 steps as 1/12 (successful events/total events)

the 12 events being:

{A->B->C , A ->B->G, A->B->A , A ->F->E, A->F->A , A ->F->G, A->G->C , A ->G->A, A->G->B , A ->G->E, A->G->F , A ->G->D, }

The 1st answer is correct, but how? What are these 18 possible paths?

答:
You have to fix your $4$th event from $A \rightarrow F \rightarrow D$ to $A \rightarrow F \rightarrow E$ obviously.

That said, your $12$ events do not have the same probability. The first $6$ on the list have probability $1/9$, the other $6$ have probability $1/18$, because they pass by $G$. You want the probability of the last event $A \rightarrow G \rightarrow D$, which can be expressed as:

$$\frac{1}{18}=\frac{\frac{1}{18}}{6\frac{1}{9}+6\frac{1}{18}} \neq \frac{1}{6+6} = \frac{1}{12}$$