问:
Translation invariant ultrafilters?
Is there an ultrafilter $\mathcal{U}$ on $\mathbb{N}$ such that $\mathcal{U}-n = \mathcal{U}$ for all $n \in \mathbb{N}$? Here, $$\mathcal{U}-n = \{ A - n: A \in \mathcal{U} \}$$ and $$A - n = \{ m \in \mathbb{N}: m+n \in A \}$$

答:
Let $E$ and $O$ be the set of even and odd naturals respectively, and note that $E-1=O$ and $O-1=E$. If $\mathcal{U}$ is an ultrafilter, either $E\in\mathcal{U}$ and $O\not\in\mathcal{U}$ or $E\not\in\mathcal{U}$ and $O\in\mathcal{U}$. Either way, there is an $X\in\mathcal{U}$ such that $X-1\not\in\mathcal{U}$, so $\mathcal{U}\not=\mathcal{U}-1$.

By a similar argument (consider "alternating blocks of size $n$") we get $\mathcal{U}\not=\mathcal{U}-n$ for every ultrafilter $\mathcal{U}$ and every nonzero $n\in\mathbb{N}$. So in fact we can never have the equality in the OP happen for even a single (nonzero) natural number.

Title: Translation invariant ultrafilters?


问:
Number Theory : $11x \equiv44\pmod{64}$ how to solve?
how to solve this $11x\equiv 44\pmod{64}$ ?

attempt:

first of all need to find the inverst of $11x\equiv 44\pmod{64}$

by Euclid algorithm then:

$64=11\cdot 5 +9$

$11=1\cdot 9 +2$

$9=2\cdot4+1$

$1=9-2\cdot4$

what is the next step for the make the inverst I do not understand ?

答:
how to solve this $11x\equiv 44\pmod{64}$ ?

first of all need to find the inverst of $11x\equiv 44\pmod{64}$

But we can cancel an invertible from both sides of a congruence without knowing the value of the inverse. By Bezout, since $\,\gcd(11,64)=1,\,$ we know that $\, 11^{-1}$ exists $\!\bmod 64,\,$ so $\rm\color{#0a0}{scaling}$ our congruence by $\,\color{#0a0}{11^{-1}}\,$ scales the coef of $\,x\,$ to be $1,\,$ effectively solving it for $\,x,\,$ explicitly

$$\:\!\color{#0a0}{11}\:\!x\:\!\equiv\:\! \color{#0a0}{11}\cdot 4\,\ \overset{\large \times\, \color{#0a0}{11^{-1}}\!\!\!\!\!\!}{\underset{\color{#c00}{\large \times\,11}}{\color{#c00}\Longleftarrow}\!\color{#0a0}\Longrightarrow}\,\ x\equiv 4\qquad\qquad$$

Via modular fractions $\,x \equiv {\large \frac{44}{11}}\equiv {\large \frac{4\:\!\cdot\:\! \color{#0a0}{11}}{\color{#0a0}{11}}}\equiv 4\:$ i.e. $\,4\color{#0a0}{(11)11^{-1}}\equiv 4,\,$ mod any $\,n\,$ coprime to  $11.\,$

Note that $\color{#0a0}{\text{scaling by $\,\color{#0a0}{11^{-1}}$}}$ is an invertible operation, with inverse being $\color{#c00}{\text{scaling by $11$}}$, so it results in an equivalent congruence (otherwise the operation might yield an inequivalent congruence with extraneous roots, so we'd need to verify that $\,x\equiv 4\,$ is actually a root). In fractional form this scaling by $\,11^{-1}$ is more commonly viewed as cancelling a common factor $11$ from top & bottom.

Recall that generally modular fractions are well-defined (uniquely exist) iff the denominator is coprime to the modulus. Thus reducing fractions by cancelling common factors  is always valid, since a factor of the denominator  remains coprime to the modulus, so it is invertible, so cancellable. Similarly all the other common laws of fractions remain valid for modular fractions as long as we restrict to fractions writable with invertible denominator $\:\!d\,$ (i.e. $\,d\,$ is coprime to the modulus).

ContentLicense: CC BY-SA 4.0


问:
Steps for Velocity given vector
I am asked to find velocity for the space curve
$$r(t) = \langle \cos(3t), 1-\sin(3t), 3\sin(3t)+\cos(3t)\rangle$$
at $t=\pi / 4$

I have differentiated each component and substituted $t$ in said components. Are those the correct steps to find the answers?

I got this as my answer:

$$\langle -3 \sqrt{2},  3 \sqrt{2} , -9 \sqrt{2} - 3 \sqrt{2} \rangle.$$

答:
The steps you stated you did are the correct ones. In particular, differentiating each component leads to

$$\frac{dr(t)}{dt} = \langle -3\sin(3t), -3\cos(3t), 9\cos(3t) - 3\sin(3t) \rangle \tag{1}\label{eq1A}$$

Next, $t = \frac{\pi}{4}$ means $3t = \frac{3\pi}{4}$, with $\sin(3t) = \frac{1}{\sqrt{2}}$ and $\cos(3t) = -\frac{1}{\sqrt{2}}$. Thus, \eqref{eq1A} gives

$$\left\langle -\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}, -\frac{9}{\sqrt{2}} - \frac{3}{\sqrt{2}} \right\rangle \tag{2}\label{eq2A}$$

However, your answer has the result in \eqref{eq2A} multiplied by $2$. Also, note that the $z$ component can be simplified to

$$-\frac{9}{\sqrt{2}} - \frac{3}{\sqrt{2}} = -6\sqrt{2} \tag{3}\label{eq3A}$$

Title: Steps for Velocity given vector


问:
Question about proof in Neukirch
I am reading the proof of Proposition 2.10 in Neukirch. Here is the result, given the classic $AKBL$ setup.

Prop. If $L|K$ separable and $A$ is a PID, then every finitely generated $B$-submodule $M\neq 0$ of $L$ is a free $A$-module of rank $[L:K]$. In particular, $B$ admits an integral basis over $A$.

Proof: Let $M\neq 0$ be a finitely generated $B$-submodule of $L$ and $\alpha_1,\ldots,\alpha_n$ a basis of $L|K$. Multiplying by an element of $A$, we may arrange the $\alpha_i$ to lie in $B$. We then have $dB\subseteq A\alpha_1+\cdots+A\alpha_n$ by Lem 2.9 ($d$ is discriminant), in particular, rank$(B)\leq[L:K]$, etc.

I am wondering how rank $(B)$ is even defined. Aren't we suppsoed to prove that $B$ is a free $A$-module? So how does this final assertion work and why is it true?

答:
I have been struggling to understand Neukirch's proof here as well, and I shall share my understanding of his proof below. I'm also a new learner on algebraic number theory, so I'm sincerely sorry if there are mistakes and misunderstandings.

Personally, I think that one should first prove a weaker result:

Proposition: Let $A$ be an integrally closed integral domain with field of fraction $K$, and let $B$ the integral closure of $A$ in a separable extension $L$ of $K$ of degree $m$. Then there exists free $A$-submodules $M$ and $M^{\prime}$ of $L$, such that $M \subset B \subset M^{\prime}$.

And in paricular,

Corollary $(\star)$: When $A$ is noetherian, $B$ is a finitely generated $A$-module, and furthermore, when $A$ is a PID, $B$ is free of rank $[L:K]$.

The corollary gives the second result (after In particular) of Proposition (2.10) in Neukirch. So it has already been enough if we are only interested in the existence of integral basis of $B$ over $A$. I shall provide a proof of the entire Proposition (2.10) in Neukirch using the above corollary.

PROOF of the quoted proposition and corollary $(\star)$: The proof of the above proposition and corollary can be found as Proposition 2.29 in Milne's note "Algebraic Number Theory", or [Zariski-Samuel] Vol.1, Ch.V., Sect. 4.

PROOF of (2.10): So to prove (2.10), Let $(\mu_i)_{i=1, \ldots, r} \in M$ be a system of generators of the $B$-module $M$. Then there is an $a \neq 0 \in A$ such that $a \mu_i \in B$ for $i=1, \ldots, r$. (This is the standard trick introduced on Page 8). So we see that $aM \subset B$. By Corollary $(\star)$, $B$ is a free $A$-module, where $A$ is a PID, hence $aM$ is a free $A$-module with $\text{rank}_A(aM) \leq [L:K]$. So $M$ is also a free $A$-module with $\text{rank}_A(aM) \leq [L:K]$.

To show that $\text{rank}_A(M) = [L:K]$, it remains to show that $\text{rank}_A(aM) \geq [L:K]$. This is discussed in the post: Incomplete Proof from Neukirch about modules and algebraic integers.

Just to restate the discussion there (Some of the arguments below are directly copied there from Hagen Knaf's answer.), let $(\alpha_i)_{i=1, \ldots, n}$ be a basis of $L|K$. Multiplying by an element of $A$, we can arrange for $\alpha_i \in B$. (Still the trick introduced on Page 8) Take an element $m \in M \setminus 0$. Then the elements $\alpha_i m\in M$, $i=1,\ldots ,n$, are linearly independent over $A$, which means that the $A$-submodule $Bm$ of $M$ has rank $n$. Hence
$$
\text{rank}_A (M) \geq  \text{rank}_A (Bm) = n = [L:K].
$$

This is what we desired.

Sorry for any possible mistakes above. And I'm still wondering if there is a good way to prove (2.10) directly, or understand what Neukirch means.

THANKS: Thanks Angina Seng for the comments in this post, and Hagen Knaf for his answer in the linked post above!

Title: Question about proof in Neukirch


问:
limit of complex numbers $\lim_{z\to 0} \frac{\bar{z}^2} {z}$
$$\lim_{z\to 0} \frac{\bar{z}^2} {z}$$

How can we calculate this limit

答:
Hint

Use the polar coordinates to express $z=re^{i\theta}$ with $r\to 0$.


问:
proof of one matrix rank inequality
I forgot how to prove the following inequality. Could anybody help?

Let $A, B, C$ be matrices of appropriate dimensions:
$$rank\left( \left[\begin{array}{rr}
A & 0\\
C & B
\end{array}\right]\right)\ge rank\left( \left[\begin{array}{rr}
A & 0\\
0 & B
\end{array}\right]\right).$$

答:
Let $c'_1,..., c'_n$ be the columns of

$$\left( \left[\begin{array}{rr}
A & 0\\
C & B
\end{array}\right]\right)$$

and $c_1,..., c_n$ be the colums of

$$\left( \left[\begin{array}{rr}
A & 0\\
0 & B
\end{array}\right]\right)$$

If $c_{i_1},..., c_{i_j}$ are linearly independent and

$$k_1c'_{i_1}+...+ k_jc'_{i_j} =0 \,.$$

then, it follows immediately by writing explicitely the columns that $k_1=..=k_j=0$.

Hence if we pick a basis in the column  space of $\left( \left[\begin{array}{rr}
A & 0\\
0 & B
\end{array}\right]\right)$, the correcponding colums are linearly independent in $\left( \left[\begin{array}{rr}
A & 0\\
C & B
\end{array}\right]\right)$, which proves the inequality.

ContentLicense: CC BY-SA 3.0

Title: proof of one matrix rank inequality


问:
Under what conditions is a zero divisor element $a$ in commutative ring $R$ nilpotent?
Suppose that $R$ is a commutative ring with identity $1$

Let $a\in R$ with $ab=0$ for some $b\ne0$.

Under what conditions $a$ must be also nilpotent?

答:
I don't think there is any better answer for this question than the more specific ones given at your other question "Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? ". The most natural conditions are mentioned there, and they are generally conditions on ideals of the ring, not the specific element $a$.

To recap some of the sufficient conditions mentioned there that make zero divisors nilpotent:

$R$ an Artinian local ring

$\{0\}$ a primary ideal.


答: Note that $| { \overline{z}^2 \over z } | = | { z^2 \over z}| = |z|$. Hence $\lim_{z \to 0} { \overline{z}^2 \over z } = \lim_{z \to 0} z = 0$.

Title: limit of complex numbers $\lim_{z\to 0} \frac{\bar{z}^2} {z}$

Body:
$$\lim_{z\to 0} \frac{\bar{z}^2} {z}$$

How can we calculate this limit


问:
Path connected sets (How to intutively find which set is path connected before attempting a formal proof )
I have been preparing for an objective masters exam and  general topology is in the syllabus. The instructor in undergrad course was not interested in teaching and so I self studied and managed to do average level understanding of topology.

Problem arises in path connected set . Proving a set  path connected by definition is not easy and questions are often asked in exam whether a set is path connected or not? But rigorious proof is not asked as I have to just mark the correct options. So, I am asking for if there is some intution .  Consider that  I am given a set and  I have to see whether it's path connected or not , then I can use that intutive method which is less rigorious and time saving also.

If yes , can you tell me by using it on this question : Let A be the following subset of $\mathbb{R}^2$ : A={$(x,y)$:${(x+1)}^2+y^2 \leq  1$} $\bigcup $ {$(x,y): y = x sin(1/x), x>0$}. (This was 1 of the question asked in that exam).

I will apply your method/explanation/intution to other asked question for practice.

I shall be really thankful for any help received!

答:
Let $E = \bigg\{ (x,y)\bigg|y=x\sin\ \frac{1}{x},\ 0<x\leq 1\bigg\}$. We have a claim that $E'=\{(0,0) \} \bigcup E$ is path-connected,
i.e. we must construct a continuous map $c : [0,1] \rightarrow
(\mathbb{R}^2,|\ |)$ where $|\ |$ is Euclidean distance s.t.
$$c(0)=(0,0),\ c(1) =(1,\sin\ 1)$$ and $c([0,1])$ is in $ E'$.

Define $c_n : [0,1]
 \rightarrow
(\mathbb{R}^2,|\ |)$ s.t. $c_n(t)=(t,t\sin\ \frac{1}{t}),\
\frac{1}{n}\leq t$, $$ c_n(s) =  \frac{s}{\frac{1}{2n}}
\bigg(\frac{1}{n},0\bigg) ,\ 0\leq s\leq \frac{1}{2n} $$ and $$ c_n(s) =
\frac{ \frac{1}{n}  -s }{\frac{1}{2n} }  \bigg(\frac{1}{n},0\bigg)
 + \frac{s- \frac{1}{2n}}{\frac{1}{2n}} \bigg(\frac{1}{n},\frac{1}{n}\sin\ n\bigg),\ \frac{1}{2n}<s<\frac{1}{n}
$$

Define $c(0)=(0,0),\ c(t)=(t,t\sin\ \frac{1}{t})$ so that
$|c_n(t)-c(t)| \leq  \frac{2\sqrt{2}}{n}
 $ for all $t$.

Hence $c$ is a uniform limit of continuous curves
$c_n$. That is $c$ is continuous.


问:
Help with the contour for this integral using residues
$$ PV \int_0^\infty \frac{dx}{\sqrt{x}(x^2-1)} $$

A keyhole contour can't be used because we have a pole in the real positive axis, isn't it?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\mrm{P.V.}\int_{0}^{\infty}{\dd x \over
\root{x}\pars{x^{2} - 1}}} =
\Re\int_{0}^{\infty}{\dd x \over
\root{x}\pars{x - 1 + \ic 0^{+}}\pars{x + 1}}
\\[5mm] = &\
-\,\Re\int_{\infty}^{0}{\ic\,\dd y \over
\root{y}\expo{\ic\pi/4}\pars{\ic y - 1 + \ic 0^{+}}
\pars{\ic y + 1}}
\\[5mm] = &\
-\,\Re\int_{0}^{\infty}{\ic\expo{-\ic\pi/4}\, \over
\root{y}\pars{1 + y^{2}}}\,\dd y =
-\,{\root{2} \over 2}\int_{0}^{\infty}{y^{-1/2} \over
1 + y^{2}}\,\dd y
\\[5mm] = &\
-\,{\root{2} \over 4}\int_{0}^{\infty}{y^{\color{red}{1/4} - 1} \over
1 + y}\,\dd y
\end{align}
Note that $\ds{{1 \over 1 + y} =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}}
\,{\pars{-y}^{k} \over k!}}$. With Ramanujan's MT:
\begin{align}
&\bbox[5px,#ffd]{\mrm{P.V.}\int_{0}^{\infty}{\dd x \over
\root{x}\pars{x^{2} - 1}}} =
-\,{\root{2} \over 4}\Gamma\pars{\color{red}{1 \over 4}}
\Gamma\pars{1 - \color{red}{1 \over 4}}
\\[5mm] = &\
-\,{\root{2} \over 4}\,{\pi \over \sin\pars{\pi/4}} =
\bbx{-\,{\pi \over 2}} \\ &
\end{align}


Title: Number Theory : $11x \equiv44\pmod{64}$ how to solve?

Body:
how to solve this $11x\equiv 44\pmod{64}$ ?

attempt:

first of all need to find the inverst of $11x\equiv 44\pmod{64}$

by Euclid algorithm then:

$64=11\cdot 5 +9$

$11=1\cdot 9 +2$

$9=2\cdot4+1$

$1=9-2\cdot4$

what is the next step for the make the inverst I do not understand ?

Body: $11x  \equiv 44 ($ mod $64 $) $=> 64|(11x-44)=> 64|11(x-4).$ As $gcd(64,11)=1$ , therefore it implies $64|(x-4)$. Therefore $64k= x-4=> x=64k+4$


问:
$\|f*g\|_q\leq \|g\|_q \|f\|_1$ and $\|f*g\|_\infty\leq \|g\|_q \|f\|_{q^{'}}$, $(1/q+1/q^{'}=1)$?
Now I'm reading the Young inequality. It says that if $f \in L^p(R)$, $g \in L^q(R)$, $1\leq p,q\leq \infty$, $1/p+1/q\geq 1$. Then how could we have the following inequalities:

$$\|f*g\|_q\leq \|g\|_q \|f\|_1$$

$$\|f*g\|_\infty\leq \|g\|_q \|f\|_{q^{'}},\quad (1/q+1/q^{'}=1)$$

$f*g$ means the convolution

The two inequalities seems not obvious, the second one is similar with the holder inequality, but actually not.

Could someone give me some hints or helpful links?
Thanks a lot!

答:
Here is the first inequality using Jensen instead of Young.

If $\|f\|_1=1$, we can use Jensen's inequality on the measure $\mathrm{d}\mu=f(x-y)\,\mathrm{d}y$ to get
$$
\left|\int f(x-y)g(y)\,\mathrm{d}y\right|^p\le\int |f(x-y)||g(y)|^p\,\mathrm{d}y
$$
Then integrate with respect to $x$ to get
$$
\|f\ast g\|_p^p\le\|f\|_1\|g\|_p^p=\|f\|_1^p\|g\|_p^p
$$
since $\|f\|_1=1$. This equation now scales nicely with $\|f\|_1$, so we can remove the restriction that $\|f\|_1=1$ to get
$$
\|f\ast g\|_p\le\|f\|_1\|g\|_p
$$

As mentioned by julien, the second inequality is simply Hölder's inequality.


问:
Reverse order of polynomial coefficients of type $\left(r-x\right)^n$
I have a simple question.

I have a polynomial defined by
$$
\left(r-x\right)^n = a_0 + a_1 x + ... + a_n x^n
$$

Is there a simple expression if I put the coefficients in reverse order?

$$
a_n + a_{n-1} x + ... + a_1 x^{n-1} a_0 x^n = ?
$$

If $r = 1$ it is easy but for other $r$ I don't know how to do it.

Edit solution:
Thanks to the general answer of paul blart math cop, the simple expression with reversed coefficient order is:

$$
x^n \left(r-\frac{1}{x}\right)^n = \left(rx-1\right)^n
$$

答: Letting $f(x)$ be a polynomial of degree $n$, the polynomial with reversed coefficients to $f$ is $x^n f(1/x)$.


问:
Minimizing $\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$ over positive reals with $a+b=1$. Why is the minimum not $18$?

If $a,b \in R^+$ such that $a+b=1$, then find the minimum value of $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$$

We can write $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}=\left(a+\frac{a+b}{a}\right)^{2}+\left(b+\frac{a+b}{b}\right)^{2}=\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2}$$
Using $Q.M\geq A.M\geq G.M$ we have
$$\sqrt{\frac{\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2}}{2}} \geq \frac{2+\frac{a}{b}+2+\frac{b}{a}}{2}=2+\frac{\frac{a}{b}+\frac{b}{a}}{2} \geqslant 2+1=3$$
So
$$\begin{array}{l}
\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2} \geqslant 18 \\
\Rightarrow \quad\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geqslant 18
\end{array}$$

But in an alternate approach i got the correct minimum as $12.5$

答:
The issue is in the first line of simplification.
$$a+\frac{a+b}{a}=a+1+\frac{b}{a}\neq 2+\frac{b}{a}.$$
If you make the correct simplification, you can continue
$$\sqrt{\frac{\left(a+1+\frac ba\right)^2+\left(b+1+\frac ab\right)^2}{2}}\geq \frac{a+b+2+\frac ab+\frac ba}{2}=\frac{3+\frac ab+\frac ba}{2}\geq \frac 52,$$
and get the correct result.


问:
Is there (or can there be) a general algorithm to solve Rubik's cubes of any dimension?
I love solving Rubik's cube (the usual 3D one). But, a lecture by Matt Parker at the Royal Institute (YouTube Link) led me to an app that can simulate a four dimensional rubik's cube. But unfortunately it was so complex, that I soon got bored as I failed to solve it.

The website to the 4D cube : http://superliminal.com/cube/cube.htm

Following which today, I also found an app that can simulate a 5D cube!!

The website to the 5D cube: http://www.gravitation3d.com/magiccube5d/

So, my two questions are :

Is there a general (non brute-force) algorithm that can be used to solve a well-scrambled cube of any dimension (even though it may not be very efficient, but yet is not a simple search over all the available space of move sequences) ? [Point modified after reading @RavenclawPrefect's answer :D ]

Mathematically, what is common in all these cubes, and "hypercubes"?

NOTE: By dimensions I mean physical dimension and not the number of rows of pieces on a face of the cube. So a four-dimension cube is a cube in x,y,z,$\delta$ dimensions where $\hat x,\hat y,\hat z,\hat \delta$ are mutually orthogonal unit vectors in the respective dimensions.

For example, here is how the aforementioned 4D cube moves:
https://miro.medium.com/max/2552/1*ga32DoV_Hc6e8t6PC1hFHw.gif

Addendum:

List of resources that may help finding an answer to this question:

Solving Rubik's cube and other permutation puzzles

答:
Here is a very simple (and very inefficient) algorithm:

The set of finite sequences of moves is countable, so we can enumerate such sequences $s_1, s_2, \ldots$ (specifically, we can do so in a computable manner). Then execute the following algorithm:

```<code>i=1
while cube is not yet solved:
  execute move sequence s_i
  if cube is solved:
    return s_i
  else
    undo move sequence s_i
  i = i+1
</code>```

Since every finite sequence of moves appears in our enumeration, this algorithm will always terminate in finite time (and in fact for a fixed puzzle with finitely many configurations, its runtime will necessarily be bounded).

For a more "reasonable" algorithm, some kind of bound needs to be put on the runtime; this one is at least exponential in the dimension of the cube, which I suspect is very suboptimal.

Edit: This answer seems to have gotten a lot of attention by virtue of being the earliest one to the question, but the other answers to this question are significantly more interesting; go upvote them instead!

Body:
Here is a very simple (and very inefficient) algorithm:

The set of finite sequences of moves is countable, so we can enumerate such sequences $s_1, s_2, \ldots$ (specifically, we can do so in a computable manner). Then execute the following algorithm:

i=1
while cube is not yet solved:
  execute move sequence s_i
  if cube is solved:
    return s_i
  else
    undo move sequence s_i
  i = i+1

Since every finite sequence of moves appears in our enumeration, this algorithm will always terminate in finite time (and in fact for a fixed puzzle with finitely many configurations, its runtime will necessarily be bounded).

For a more "reasonable" algorithm, some kind of bound needs to be put on the runtime; this one is at least exponential in the dimension of the cube, which I suspect is very suboptimal.

Edit: This answer seems to have gotten a lot of attention by virtue of being the earliest one to the question, but the other answers to this question are significantly more interesting; go upvote them instead!


问:
Integration(Improper Integration)
$$
\mbox{Integrate}\quad
\int_{0}^{\infty}\frac{\mathrm{d}x}
{\left(x + 2\right)\left(x^{2} + 1\right)}.
$$

My answer is infinity, because:

$$
\frac{1}{5}\,\ln\left(b + 2\right)-
\frac{1}{10}\,\left(b^{2} + 1\right) +
\frac{2}{5\arctan\left(b\right)} -
\frac{1}{5}\,\ln\left(2\right)
$$
when $b$ gets close to infinity. Is this correct $?$.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\large\ln}$ is its Principal Branch-Cut.
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\dd x
\over \pars{x + 2}\pars{x^{2} + 1}}} =
-\,\Im\int_{0}^{\infty}{\dd x
\over \pars{x + 2}\pars{x + \ic}}
\\[5mm] = &\
-\,\Im\int_{0}^{\infty}\pars{%
{1 \over x + 2} - {1 \over x + \ic}}{1 \over \ic - 2}\,\dd x
\\[5mm] = &\
{1 \over 5}\,\Im\bracks{\pars{2 + \ic}
\ln\pars{x + 2 \over x + \ic}}_{\,0}^{\,\to\ \infty} =
{1 \over 5}\,\Im\bracks{\pars{2 + \ic}
\ln\pars{\ic \over 2}}
\\[5mm] = &\
{1 \over 5}\,\Im\braces{\pars{2 + \ic}
\bracks{-\ln\pars{2} + {\pi \over 2}\,\ic}} =
\bbx{\pi - \ln\pars{2} \over 5} \approx 0.4897 \\ &
\end{align}

Title: Integration(Improper Integration)


问:
How to prove that $|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1})$ when $x,y \geq 0$ and $p \geq 1$?
For $x \geq 0$, $y \geq 0$, prove that

$$
|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1}).
$$

I thought it would be simple but I messed everything up. Here are my attempts.

Fixing $x > y \geq 0$, I thought about the function

$$
h(p)=x^p-y^p-p(x-y)(x^{p-1}+y^{p-1})
$$

Then $h(1)=0$ and I wanted to prove that $h(p) \leq 0$ when $p \geq 1$ by discussing its derivative but the derivative is messy offering no way out.

Also I thought about after fixing $y \geq 0$, consider the function

$$
\varphi(x)=x^p-y^p-p(x-y)(x^{p-1}+y^{p-1})
$$

where $\varphi(y)=0$ and I want to prove that $\varphi(x) \leq 0$ when $x \geq y$ yet again it's a messy way.

I also thought about restricting $p$ to $\mathbb{N}$ or $\mathbb{Q}$ using decomposition like

$$
(x^p-y^p)=(x-y)(x^{p-1}+x^{p-1}y+x^{p-2}y^2+\cdots+y^{p-1})
$$

but the number of terms does not match.

But I do believe this should be a easy question with some special background. I must have missed something critical. Any hint/solution appreciated!

答:
Wlog $x>y$. Then with $f(t):=x^p$, we have from the Mean Value Theorem that there exists some $\xi$ with $x>\xi>y$ such that
$$\frac{x^p-y^p}{x-y}=\frac{f(x)-f(y)}{x-y}=f'(\xi) =p\xi^{p-1}\le p\max\{x^{p-1},y^{p-1}\}<p(x^{p-1}+y^{p-1}).$$


问:
If you had a large group of people what would be the fastest way for the people to count themselves?
(I realize this has a computer-sciency feel but I am pretty sure this is more about math than it is about computers. Maybe it is about information or game theory?)

I assume that there would be some partitioning and each group could count themselves in parallel with other groups and then perhaps pass the count along when done. You can eyeball the crowd but there is no explicit information at all.

What's more, the crowd in not in an enclosed area where by roughly calculating the area that the crowd packs and how much space each person occupies you could get an estimate quickly. On the other hand, the group could agree to crowd together.

The group could line up and count off and that would be probably pretty accurate; perhaps the line could try to do this in parallel by partitioning the line.

But all these shufflings and groupings would take time in themselves.

I think trying to form a square or rectangle might be the fastest way to get a count that is within 10% or so of the true number.

Importantly, there is no vantage point from which any member of the crowd is allowed to look and use techniques as listed here: https://en.wikipedia.org/wiki/Crowd_counting#:~:text=The%20most%20common%20technique%20for,the%20number%20of%20sections%20occupied.

It is a completely flat plane. Everyone is of the same height and you can't make a human pyramid, etc. to give you the vantage point. Only the people at the edges know they are at the edges but they can of course share that information and everyone cooperates as much as possible.

You have to take into account that there are no leader chosen beforehand although they can elect leaders but if there are such leaders, this has to be conveyed.

Nobody has covid. There is no internecine fighting or violence of any kind but there can be confusion about, for example, who the leader or leaders are. People take up exactly the same amount of space. No fat or thin people -- and not that it matters but everyone is equally attractive, sort of average looking.

答:
Use Jacob's method.

According to Wikipedia:

[You] [divide] the area occupied by a crowd into sections, determining an average number of people in each section, and multiplying by the number of sections occupied.


问:
Why is noncentral t-distribution for sample size determination for small sample?
For sample size determination for small sample, population distribution of which is normal, the test statistic $$T = \frac{\bar{x}-\mu_0}{S/\sqrt{n}}, ~~~\text{where $\mu_0$ is null value}$$

is considered under the assumption of the alternative hypothesis $\mu = \mu' \gt \mu_0$.
Since $T$ follows noncentral $t$ distribution, chart or table is used to determine sample size $n$ with desired type II error.

The above is typical explanation available in statistics text books.

Now I'm curious what's wrong with the following reasoning:

If we assume that the alternative hypothesis $μ=μ′$ is correct $T = \frac{\bar{x} - \mu'}{S/\sqrt{n}}$ should follow $t$-distribution by the same argument that $T = \frac{\bar{x} - \mu_0}{S/\sqrt{n}}$ follows $t$-distribution when  $μ=μ_0$. Then we can proceed on to calculate

$$
Pr(\frac{\bar{x}-\mu'}{S/\sqrt{n}} \leq \frac{\mu_0 + t_{\alpha, n-1}s/\sqrt{n} - \mu'}{s/\sqrt{n}}) = \beta(\mu')~~~\text{(*)}
$$

to determine sample size n that satisfies the required $\beta$.
noncentral t-distribution cannot be involved at all.

Specifically, if my assumption that $(\bar{x}-\mu')/(s/\sqrt{n})$ follows t distribution is correct, we can simply set:

$$
\frac{\mu_0 + t_{\alpha, n-1}s/\sqrt{n} - \mu'}{s/\sqrt{n}} = t_{\beta(\mu')}
$$

and solve for $n$. No need to summon noncentral t distribution, no need to use $\beta$ curve to find $n$

=================================================================
With help of @BruceET, I found my silly mistake.

The equation to determine a rejection region is

$$
Pr(\frac{\bar{x} - \mu_0}{S/\sqrt{n}} > t_{\alpha, n-1}) = \alpha
$$

From this I wrongly thought the rejection region in terms of $\bar{x}$ is

$$
\bar{x} > \mu_0 + t_{\alpha, n-1}\frac{s}{\sqrt{n}}
$$

This is wrong because $s$ itself is a random variable and therefore cannot be included in the bounds. If it were, I could have argued the eq.(*) could be used to calculate power for the alternative hypothesis.

答:
Let $X_1, \dots, X_n$ be a random sample from $\mathsf{Norm}(\mu, \sigma).$ Suppose you want to test $H_0: \mu=10$ against $H_a: \mu > 10$ at the 5% level of significance when $n = 16.$

Then the t statistic for this test is
$T_0 = \frac{\bar X - 10}{S/\sqrt{n}}$ and $H_0$ is rejected if $T_0 \ge t^* = 1.753,$ where $t^*$ cuts
probability 5% from the upper tail of Student's t
distribution with $\nu =n-1 = 15$ degrees of freedom:
$\mathsf{T}(\nu=15).$ Computation in R.

```<code>qt(.95, 15)
[1] 1.75305
</code>```

The power $\pi(12) = P(\mathrm{Rej}\, H_0\, |\, \mu = 12)$ against the specific alternative $\mu_a = 12$ is

$$\pi(\mu_a) = \pi(12) = P\left(\frac{\bar X - \mu_0}{S/\sqrt{n}}\ge t^*\,|\mu_a=12\right)
= P\left(\frac{\bar X - \mu_a + (\mu_a - \mu_0)}{S/\sqrt{n}}\ge t^*\,|\mu_a=12\right)\\
= P\left(\frac{\bar X - \mu_a + 2}{S/\sqrt{n}}\ge t^*\,|\mu_a=12\right)
= P\left( \frac{ Z + \delta }{ \sqrt{V/\nu} } \ge t^* \right),$$
where $\delta = \sqrt{n}(\mu_a-\mu_0)/\sigma = 2\sqrt{n}/\sigma,\,$
$Z \sim\mathsf{Norm}(0,1),$ and $V = \nu S^2/\sigma^2
\sim \mathsf{Chisq}(\nu).$

That is, by definition,  $\frac{ Z + \delta }{ \sqrt{V/\nu}}\sim \mathsf{T}(\nu, \delta),$ Student's noncentral t distribution with $\nu$ degrees of freedom and noncentralty parameter $\delta.$

In particular, for our example, $\delta = 2\sqrt{16}/\sigma.$
For $\sigma = 2,$ we have $\delta = 4$ and $\pi(12) = 0.985.$

```<code>1 - pt(qt(.95, 15), 15, 4)
[1] 0.9848477
</code>```

The following simulation in R gives essentially the same power. With a million iterations one
can expect three-place accuracy.

```<code>set.seed(1128)
pv = replicate(10^6,  t.test(rnorm(16, 12, 2),
               mu=10, alt="g")$p.val)
mean(pv <= .05)
[1] 0.984809
</code>```

Ref: Except for notation, parts of the above parallel Sect 2.3 of Bain & Englehardt (1992), Intro. to Probability and Math. Stat., Duxbury, p400.

Body:
Let $X_1, \dots, X_n$ be a random sample from $\mathsf{Norm}(\mu, \sigma).$ Suppose you want to test $H_0: \mu=10$ against $H_a: \mu > 10$ at the 5% level of significance when $n = 16.$

Then the t statistic for this test is
$T_0 = \frac{\bar X - 10}{S/\sqrt{n}}$ and $H_0$ is rejected if $T_0 \ge t^* = 1.753,$ where $t^*$ cuts
probability 5% from the upper tail of Student's t
distribution with $\nu =n-1 = 15$ degrees of freedom:
$\mathsf{T}(\nu=15).$ Computation in R.

qt(.95, 15)
[1] 1.75305

The power $\pi(12) = P(\mathrm{Rej}\, H_0\, |\, \mu = 12)$ against the specific alternative $\mu_a = 12$ is

$$\pi(\mu_a) = \pi(12) = P\left(\frac{\bar X - \mu_0}{S/\sqrt{n}}\ge t^*\,|\mu_a=12\right)
= P\left(\frac{\bar X - \mu_a + (\mu_a - \mu_0)}{S/\sqrt{n}}\ge t^*\,|\mu_a=12\right)\\
= P\left(\frac{\bar X - \mu_a + 2}{S/\sqrt{n}}\ge t^*\,|\mu_a=12\right)
= P\left( \frac{ Z + \delta }{ \sqrt{V/\nu} } \ge t^* \right),$$
where $\delta = \sqrt{n}(\mu_a-\mu_0)/\sigma = 2\sqrt{n}/\sigma,\,$
$Z \sim\mathsf{Norm}(0,1),$ and $V = \nu S^2/\sigma^2
\sim \mathsf{Chisq}(\nu).$

That is, by definition,  $\frac{ Z + \delta }{ \sqrt{V/\nu}}\sim \mathsf{T}(\nu, \delta),$ Student's noncentral t distribution with $\nu$ degrees of freedom and noncentralty parameter $\delta.$

In particular, for our example, $\delta = 2\sqrt{16}/\sigma.$
For $\sigma = 2,$ we have $\delta = 4$ and $\pi(12) = 0.985.$

1 - pt(qt(.95, 15), 15, 4)
[1] 0.9848477

The following simulation in R gives essentially the same power. With a million iterations one
can expect three-place accuracy.

set.seed(1128)
pv = replicate(10^6,  t.test(rnorm(16, 12, 2),
               mu=10, alt="g")$p.val)
mean(pv <= .05)
[1] 0.984809

Ref: Except for notation, parts of the above parallel Sect 2.3 of Bain & Englehardt (1992), Intro. to Probability and Math. Stat., Duxbury, p400.


问:
What can be said about Radius of Convergence?
If the following complex series converges for all positive real values of $z$, what can be said about the radius of convergence ? Why ?

$\sum_{n=0}^{\infty} a_n(z-z_0)^n$

答:
Note that we have the a priori information about power series:

Given any power series, its radius of convergence $R \in [0, \infty]$ is such that the power series

converges for all $z$ with $|z - z_0| < R$,

diverges for all $z$ with $|z - z_0| > R.$

Now, if the radius $R$ is finite in this case, then we can pick a positive real $z$ outside the the circle $|z - z_0| \le R$ to get a contradiction. For example, note that $z = |z_0| + R + 43$ is one such number.

To elaborate: Assume that the radius is $R < \infty$. Then, the number $z = |z_0| + R + 43$ is a positive real and you have $|z - z_0| > R$ and thus, the power series must diverge at $z$ (since it's outside the radius of convergence) but we are given that it converges at $z$ (since $z$ is a positive real number). A contradiction.


问:
Does preadditive category require locally smallness?
In general, people define a category $\mathscr{A}$ to be preadditive if for every $A,A'\in\mathscr{A}$ the collection $\mathscr{A}(A,A')$ has an abelian group structure. However, definitions in numerous sources online (like wiki, nlab, stacks project) all seems to not require the category to be $\textit{locally small}$. But if $\mathscr{A}$ is not locally small, then some hom-sets are not necessary set-worth, so it makes no sense to give them an abelian $\textit{group}$ structure. So do people just loosen their terminology to say a collection is an abelian group even if the collection is not a set?

答:
The standard convention is that categories are locally small by default. I don't understand your claim about online sources; in fact the Stacks project definition of a category (tag 0013) requires that categories be small, except for a specific collection of exceptions, which I personally find bizarre.

In any case an abelian group must have an underlying set so a preadditive category must in particular be locally small.


问:
Prove that $\lim _{n\rightarrow \infty }\dfrac{n^{3}+1}{n^{4}+4}=0$
Prove that $\lim _{n\rightarrow \infty }\dfrac{n^{3}+1}{n^{4}+4}=0$ . I'll write the question without epsilon because I write definition on my paper. I just stuck where I wrote.

$\dfrac{n^{3}+1}{n^{4}+4} <\dfrac{n^{3}+1}{n^{4}} <\dfrac{n^{3}+n^{2}}{n^{4}} <\dfrac{n+1}{n^2}\leq\dfrac{\left| n\right| +\left| n\right| }{|n|^{2}}=2\cdot \left| \dfrac{n}{n^{2}}\right| $ Is it okay? Can I make such an inequality or are there any solution? Thanks.

答:
Almost you are reached out...

Let $\epsilon>0$. Let $N=\frac{\epsilon}{2}$, Therefore for all $n>N$.

\begin{equation}
\left| \frac{n^3+1}{n^4+1}-0 \right|= \left| \frac{n^3+1}{n^4+1}\right|\leq 2\frac{1}{n}<2\frac{1}{N}=\epsilon.
\end{equation}


问:
Evaluation of integral $\int_{0}^{\infty}\frac{\sin x}{x\left ( 1+x^2 \right )^2}\,{\rm d}x$
I'm trying to evaluate the following integral:

$$\mathcal{J}=\int_{0}^{\infty}\frac{\sin x}{x\left ( 1+x^2 \right )^2}\,{\rm d}x$$

Well there are $3$ poles , one lying on the real line the other on the upper half plane and the other on the lower half plane. The residue at $z=i$ is $\displaystyle -\frac{3}{4e}$ .
I'm integrating on a contour that looks like a semicircle on the upper half plane and has a branch about the origin .

Well I considered the function $\displaystyle f(z)=\frac{e^{iz}+1}{z\left ( z+1 \right )^2}$ which is clearly analytic expect for the poles. Hence if $\gamma$ denotes the contour , then:

$$\oint_{\gamma}f(z)\,{\rm d}z=\oint_{\gamma}\frac{e^{iz}+1}{z\left ( z^2+1 \right )^2}\,{\rm d}z =\oint_{\gamma}\frac{e^{iz}}{z\left ( z^2+1 \right )^2}\,{\rm d}z+\oint_{\gamma}\frac{{\rm d}z}{z\left ( z^2+1 \right )^2}=-2\pi i \frac{3}{4e}+2\pi i = 2\pi i \left ( 1-\frac{3}{4e} \right )=i \left (2\pi - \frac{3\pi}{2e} \right )$$

Hmm... applying the classical method I get that:

$$\int_{-\infty}^{\infty}f(x)\,{\rm d}x =2\pi - \frac{3\pi}{2e}$$

which is almost correct except for that $2$ in front of $\pi$. Where I have gone wrong?

P.S: I used the very obvious that the integrand is even.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\sin\pars{x} \over
x\pars{1 + x^{2}}^{2}}\,\dd x} =
{1 \over 2}\,\Im\int_{-\infty}^{\infty}{\expo{\ic x} - 1 \over
x\pars{x - \ic}^{2}\pars{x + \ic}^{2}}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\Im\braces{2\pi\ic\,\lim_{x\ \to\ \ic}\,\,\,
\totald{}{x}\bracks{\expo{\ic x} - 1 \over
x\pars{x + \ic}^{2}}}
\\[5mm] = &\
\bbx{{\pi \over 2} - {3\pi \over 4\expo{}}} \approx 0.7040 \\ &
\end{align}


问:
Proving Properties of orthogonal Matrix
I was given a task whereby its defined that a $n \times n$ matrix, A, is orthogonal if $\langle A\vec{u},A\vec{v}\rangle$ = $\langle \vec{u}$,$\vec{v}\rangle$ and i have also been given the property that

$\langle A\vec{u},\vec{v}\rangle$ = $\langle \vec{u},A^T\vec{v}\rangle$ .

Now if $A$ is an orthogonal matrice how do i prove that $A^TA = I_n$

I know how to prove it using the property that an orthogonal matrices has column vectors that are orthonormal. However we were told to prove this property in the later part so I don't think we should use that in this question. So if not using this property, I'm completely lost in proving this and I would appreciate any help at all.

答:
For some matrix $C$, we know that if $u^T C v = 0$ for all vectors $u$ and $v$, then $C = O$.
$$
\left
=
u^T A^T A v = u^T v
$$
or
$$
u^T (A^T A - I) v = 0
$$
for all vectors $u$, $v$.

Title: Proving Properties of orthogonal Matrix

Body:
For some matrix $C$, we know that if $u^T C v = 0$ for all vectors $u$ and $v$, then $C = O$.
$$
\left<A u, Av\right>
=
u^T A^T A v = u^T v
$$
or
$$
u^T (A^T A - I) v = 0
$$
for all vectors $u$, $v$.


答: If $A$ is orthogonal, then, for any two vectors $\vec u$ and $\vec v$, you have $\left\langle A\vec u,A\vec v\right\rangle=\left\langle\vec u,\vec v\right\rangle$. But you also have $\left\langle A\vec u,A\vec v\right\rangle=\left\langle\vec u,A^\top A\vec v\right\rangle$. So, for any two vectors $\vec u$ and $\vec v$,$$\left\langle\vec u,A^\top A\vec v\right\rangle=\left\langle\vec u,\vec v\right\rangle.$$In other words,$$\left\langle\vec u,\left(A^\top A-\operatorname{Id}\right)\vec v\right\rangle=0.$$In particular,$$\left\|\left(A^\top A-\operatorname{Id}\right)\vec v\right\|^2=\left\langle\left(A^\top A-\operatorname{Id}\right)\vec v,\left(A^\top A-\operatorname{Id}\right)\vec v\right\rangle=0.$$So, for every vector $\vec v$, $\left(A^\top A-\operatorname{Id}\right)\vec v=\vec0$; in other words, $A^\top A\vec v=\vec v$.

Title: Proving Properties of orthogonal Matrix

Body:
I was given a task whereby its defined that a $n \times n$ matrix, A, is orthogonal if $\langle A\vec{u},A\vec{v}\rangle$ = $\langle \vec{u}$,$\vec{v}\rangle$ and i have also been given the property that

$\langle A\vec{u},\vec{v}\rangle$ = $\langle \vec{u},A^T\vec{v}\rangle$ .

Now if $A$ is an orthogonal matrice how do i prove that $A^TA = I_n$

I know how to prove it using the property that an orthogonal matrices has column vectors that are orthonormal. However we were told to prove this property in the later part so I don't think we should use that in this question. So if not using this property, I'm completely lost in proving this and I would appreciate any help at all.


答:
WLOG $x\ge y$

let $t=\frac{x}{y}\ge 1$

we have to prove $$f(t)=(p-1)t^p-pt^{p-1}+pt-p+1\ge 0$$ notice $$f'(t)=p(p-1)(t^{p-1}-t^{p-2})+p>0$$ thus $f(t)$ is increasing also $f(1)=0$ hance $f(t)\ge 0$ for all $x,y$ in domain given (because the case $x\le y$ can be proved analogously)

Title: How to prove that $|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1})$ when $x,y \geq 0$ and $p \geq 1$?

Body:
For $x \geq 0$, $y \geq 0$, prove that

$$
|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1}).
$$

I thought it would be simple but I messed everything up. Here are my attempts.

Fixing $x > y \geq 0$, I thought about the function

$$
h(p)=x^p-y^p-p(x-y)(x^{p-1}+y^{p-1})
$$

Then $h(1)=0$ and I wanted to prove that $h(p) \leq 0$ when $p \geq 1$ by discussing its derivative but the derivative is messy offering no way out.

Also I thought about after fixing $y \geq 0$, consider the function

$$
\varphi(x)=x^p-y^p-p(x-y)(x^{p-1}+y^{p-1})
$$

where $\varphi(y)=0$ and I want to prove that $\varphi(x) \leq 0$ when $x \geq y$ yet again it's a messy way.

I also thought about restricting $p$ to $\mathbb{N}$ or $\mathbb{Q}$ using decomposition like

$$
(x^p-y^p)=(x-y)(x^{p-1}+x^{p-1}y+x^{p-2}y^2+\cdots+y^{p-1})
$$

but the number of terms does not match.

But I do believe this should be a easy question with some special background. I must have missed something critical. Any hint/solution appreciated!


问:
Conformal map from upper-half plane to upper-half plane
I am trying to solve a below problem.

Problem

Let $f$ be a conformal map from upper-half plane to upper-half plane and $f(i)=i$. Show that there exists  $\theta \in \mathbb{R}$ such that $f(z) = \frac{(cos{\theta})z + (sin{\theta})}{(-sin{\theta})z + (cos{\theta})}$.

By Schwarz lemmma and Cayley transformation $T(z)=\frac{z-i}{z+i}$, $|T\circ f\circ T^{-1} (z)| = |z|$. It seems that the lemma and the transformation is useful to solve this problem, however I cannot find the $\theta$. How can I find the $\theta$ ? Thank you.

答: Hint: Let $\phi (z)=\frac {i(i-z)} {i+z}$.  $g(z)=\phi^{-1}(f(\phi(z)))$ defines a conformal equivalence of the open unit disk such that $g(0)=0$. All conformal equivalences of the unit disk have the form $e^{i\theta }\frac {z-a} {1-\overline  a z}$ and among them the only one's vanishing at $0$ are of the form $z \to e^{i\theta} z$. [This is proved in Rudin's RCA]. Hence  $g(z)=e^{i\theta} z$ and you can write down $f(z)=\phi(g(\phi^{-1}(z)))$ from this.


问:
Addition formulas for Jacobi amplitude function
Are there any known summation formulas for the Jacobi amplitude function? I need a formula like $\mathrm{am}(t+x)=\mathrm{am}(t) + f(x)$. I have plotted some graphs and it seems that $f(x)$ is periodic but I wasn't able to figure out what this function is.

Maybe someone can give me references?

答:
Actually there is the following formula:
$$ \operatorname{am}(u+v)=x + y , $$
where
$$\tan x = \frac{\operatorname{sn} u\,\operatorname{dn} v}{\operatorname{cn} u},$$
$$\tan y = \frac{\operatorname{sn} v\,\operatorname{dn} u}{\operatorname{cn} v}.$$
One has to choose carefully, however, which determinations of the arctan to use.


问:
NP, NP-complete and Reducibility
Suppose there is polynomial time reduce from problem $A$ to $B$.

fact $1)$ if problem $B$ is $NP$-hard then Problem $A$ is $NP$-complete.

fact $2)$ if problem $A$ is $NP$-complete then Problem $B$ is $NP$-complete.

fact $3)$ problem $A$ as not hard as $B$.

Can I say fact $(3)$ is true and fact $(1)$, $(2)$ is false because:

the fact $1$ is false because if problem $B$ is $NP$-hard then $A$ can be $P$, or $NP$ or $NP$-complete?

the fact $2$ is false because if problem $A$ is $NP$-complete then problem $B$ can be $NP$-complete or $NP$-hard?

Is my reasoning correct?

答:
For 1), let both $A$ and $B$ be halting problem. Then $A$ is clearly reducible to $B$ and $B$ is $NP$-hard (any computable problem can be reduced to halting problem in polynomial time), but $A$ isn't $NP$-complete (as it's not in $NP$).

For 2), let $A$ be any $NP$-complete problem (for example, 3-SAT) and let $B$ be halting problem. Then again $A$ is reducible to $B$, but $B$ isn't $NP$-complete.

For 3) - it's unclear what "problem $A$ as not hard as $B$" means.

Title: NP, NP-complete and Reducibility


问:
What's the intuition behind this equality involving combinatorics?
What is the intuition behind

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

? I can't grasp why picking a group of $k$ out of $n$ bijects to first picking a group of $k-1$ out of $n-1$ and then a group of $k$ out of $n-1$.

答:
We have a group of $n$ people, one of whom is John. We want to pick a committee of $k$ people. By definition this can be done in $\binom{n}{k}$ ways.

There are $\binom{n-1}{k}$ committees of $k$ that don't include John, for we can choose any $k$ of the people other than John. And there are $\binom{n-1}{k-1}$ committees of $k$ that do include John, for we can choose any $k-1$ people to join John.

Note that automatically a committee that doesn't include John is different from a committee that includes John. So we have divided the $\binom{n}{k}$ possible committees into two groups, one of which has $\binom{n-1}{k}$ elements, and the other of which has $\binom{n-1}{k-1}$ elements.


问:
Integration of $(1 - \cos\theta)^n$ where $n$ is a positive integer
I came across this integral during a derivation I'm trying to figure out.

$$
f(N) = \int_0^\pi {\sum_{n=1}^N (1-\cos\theta)^n \, d\theta},\quad N \in \mathbb{N}
$$

Is there any sort of a general formula to this function for a given positive integer value of $N$?

I would appreciate any hints or some direction on the matter too. I've tried basic substitution of variables, but only got so far. The power $n$ makes it quite tricky.

答:
Using $\cos\theta=1-2\sin^2(\theta/2)$ and the beta function identity, we get

\begin{align*}
\int_{0}^{\pi} (1-\cos\theta)^n \, \mathrm{d}\theta
&= \int_{0}^{\pi} 2^n \sin^{2n}(\theta/2) \, \mathrm{d}\theta
= 2^{n+1} \int_{0}^{\pi/2} \sin^{2n}\varphi \, \mathrm{d}\varphi \\
&= 2^{n} B\left(\tfrac{1}{2},n+\tfrac{1}{2}\right)
= 2^{n} \pi \prod_{k=1}^{n} \frac{k-\frac{1}{2}}{k} \\
&= \frac{\pi}{2^n} \binom{2n}{n}.
\end{align*}

Alternatively, we may also write

$$ \int_{0}^{\pi} (1-\cos\theta)^n \, \mathrm{d}\theta = (-1)^n \pi 2^n \binom{-1/2}{n}. $$


问:
How to solve this square root inequality?
I was trying to solve an inequality with a square root (to practice for admission test),
this is the inequality:

```<code>                       7x + 2 < sqrt(8x^2 + x - 9) + x
</code>```

Here's how I solved it:
1st, I've moved the square root from the right side to the left side, to flip the sign of the inequality and the sign of each member:

```<code>                        sqrt(8x^2 + x - 9) > -x + 7x + 2
</code>```

2nd, now I've split the solutions into two different cases:

first system of inequalities:

8x^2 + x - 9 >= 0

-x + 7x + 2 > 0

8x^2 + x - 9 > (-x + 7x + 2)^2

second system:

-x + 7x + 2 < 0

3rd, I've solved each inequalities:
first system of inequalities:
8x^2 + x - 9 >= 0  -->    x < -9/8 and x < 1

-x + 7x + 2 > 0    -->    x < -1/3

8x^2 + x - 9 > (-x + 7x + 2)^2 --> 36x^2 + 24x + 4 > 0 --> Doesn't have real solutions (delta is negative) but since the inequality has "greater sign", then I can say it accepts every real solutions.

second system:
-x + 7x + 2 < 0 --> x > -1/3

4th, now I've to find the common solutions in the number line:
my result is only the empty set of possible solutions.

This exercise should give "x <= -9/8".

if you know a quickest method to solve these kind of exercises, let me know because the day of the official test I'll have only 50 minutes.

EDIT (solved):
it's possible to solve this kind of inequalities using another method.

find the domain of the square root (in this case x<-9/8 or x>1).

test by plugging arbitrary values inside the inequality (in my case, x = 2)

Let's see what happens. (in my case, 2 isn't a solution because 16 isn't less than 5.

答:
When you inspect the two cases, you are saying that the condition described in the problem is the logical disjunction of the conditions coming from each case, not their logical conjunction. Therefore you then have to consider, as solution set, the union of their solution sets (which correponds to the points that satisfy at least one of the two conditions).

Which, on a side note, is a reason why the condition $8x^2+x-9\ge0$ must also be added (this time by logical conjunction) to your conditions for case 2.
Also, in case 1 it should be $-x+7x+2\ge 0$ (or whatever the LHS is), not $-x+7x+2>0$.

There might be some "bonus" mistakes of algebra in your work, as Barry was pointing out. I did not look for them, and I don't think the discussion is ripe for addressing them.

Assuming that the mistakes are yours and not from your book, I would say that the method you are being taught for transforming equations and inequations with square roots in systems of equations and inequations is generally fine for problems of this ilk, and personally I think it's best.


答:
You don't pick a group of $k-1$ out of $n-1$ and a group of $k$ out of $n-1$; this would correspond to multiplication. You should do one or the other.

The key is to pick one of your $n$ objects as being special. Then any collection of $k$ objects from the $n$ either consists of $k-1$ objects from the $n-1$ non-special objects, plus the special one, or it consists entirely of $k$ non-special objects. There are $\binom{n-1}{k}$ collections of $k$ non-special objects, and $\binom{n-1}{k-1}$ collections of $k$ objects including the special one.

Title: What's the intuition behind this equality involving combinatorics?

Body:
What is the intuition behind

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

? I can't grasp why picking a group of $k$ out of $n$ bijects to first picking a group of $k-1$ out of $n-1$ and then a group of $k$ out of $n-1$.


问:
What is the line segment joining midpoints of opposite sides of a quadrilateral called?
I'm trying to specify a formula in terms of the line segment joining midpoints of opposite sides of a quadrilateral? Is there an accurate name for them?</p>

答:
Yes. They are called bimedians.

(Follow the link for more details.)


问:
Find sequences $a_n$ that converges to 1 and $b_n$ that converges to $\infty$ such that $a_n^{b_n}$ converges to 5
I stuck in this question for a few hours now.
Can't understand how to find one and all my guesses failed.

The Question :

Find sequence $a_n$ that converges to 1 (not constant sequence).

Find sequence $b_n$ that converges to $\infty$.

such that $a_n^{b_n}$ converges to $5$.

I found such sequences that converges to $2e$ (5.4) but it's not exactly 5..

Thank you very much and have a nice day!

答:
Hint

By noting $e=\lim_{n\to\infty} (1+{1\over n})^n$, start from

$$
a_n={1+{\ln 5\over n}}\quad,\quad b_n=n
$$

A simpler one:

$$
a_n=5^{1\over n}\quad,\quad b_n=n
$$


答: Verify that $a_n = 1 + \dfrac{\ln(5)}{n}$ and $b_n = n$ does the trick.

Title: Find sequences $a_n$ that converges to 1 and $b_n$ that converges to $\infty$ such that $a_n^{b_n}$ converges to 5

Body:
I stuck in this question for a few hours now.
Can't understand how to find one and all my guesses failed.

The Question :

Find sequence $a_n$ that converges to 1 (not constant sequence).

Find sequence $b_n$ that converges to $\infty$.

such that $a_n^{b_n}$ converges to $5$.

I found such sequences that converges to $2e$ (5.4) but it's not exactly 5..

Thank you very much and have a nice day!


问:
Show that $\gcd(a_{n+1},a_n) > a_{n-1}$ implies $a_n \geq 2^n$
Let $(a_{n})$ be an infinite sequence of positive integers such that $  \gcd(a_{n+1},a_{n}) > a_{n-1} $ for all $ n\geq 1$. How do I prove that $a_{n} \geq 2^{n}$?</p>

答:
This question has four deleted answers by people who usually know what they're doing, so I might be stepping into a minefield here, but I think this is a valid proof:

First, note that the sequence has to be strictly increasing, since $a_n\ge \gcd(a_{n+1},a_n)>a_{n-1}$.

Now consider two successive terms $a_{n-1}$ and $a_n$ that differ by a factor of two or less: $a_n\le2a_{n-1}$. Then $\gcd(a_{n+1},a_n)=a_n$, since this has to be $a_n$ divided by an integer and $a_n$ divided by $2$ is already too small. Thus, in this case $a_{n+1}$ is a multiple of $a_n$. If $a_{n+1}=2a_n$, the same argument applies to $n+1$, and so on, either to infinity or until we encounter at least a three-fold multiple.

Thus, the sequence of ratios $a_{n+1}/a_n$ exhibits the following pattern: Every ratio less than or equal to $2$ may be followed by a string of $2$s; if this ends, it is followed by an integer greater than $2$; then there may follow further ratios greater than $2$ (not necessarily integers), to infinity or until we encounter the next ratio less than or equal to $2$, and then the pattern repeats.

This suggests an induction over sections of ratios, each beginning with an integer ratio, containing an integer ratio greater than $2$ and ending in a ratio less than or equal to $2$. If the inequality $a_n \geq 2^n$ ever fails, it will necessarily have to fail at the start of one of these sections. Thus, if we can show that it holds at the start of the first such section and that the product of the ratios in each section amounts at least to the corresponding number of factors of $2$, it will follow that the inequality holds for all $n$. (If there is a last "section" that goes to infinity without ever ending in a ratio less than or equal to $2$, that can't break the inequality.)

To get the induction off the ground, consider the two cases $a_0=1$ and $a_0\neq1$. (I'm assuming the indices start at $0$, since the conditions refer to $a_{n-1}$ for $n=1$.) If $a_0=1$, the first ratio is already an integer, so the first section starts right away and the inequality is satisfied, since $a_0=1\ge2^0$. If $a_0\neq1$, then $a_0\ge2=2^1$, so we're one factor of $2$ "ahead"; subsequent ratios greater than $2$ will preserve this excess, and the first ratio less than or equal to $2$ can't fully use it up since that ratio must be greater than $1$; thus in this case, too, the inequality will be satisfied at the start of the first section.

Now for each section we want to show that the product of the ratios in that section amounts at least to as many factors of $2$. This will necessarily be the case if in multiplying all the ratios except for the low ratio at the end of the section we accumulate at least an excess factor of $2$, since the low ratio must be greater than $1$.

We can ignore any ratios of $2$ at the beginning of the section, since they don't change the balance. Then we encounter an integer ratio greater than $2$. If this is greater or equal to $4$, we're done, since we then already have the required excess factor of $2$. So the only interesting case is a ratio of $3$.

So assume the first ratio in the section after any initial twos is $a_{k+1}/a_k=3$. That already gives us an excess factor of $\frac{3}{2}$. Now $\gcd(a_{k+2},a_{k+1})$ must be a divisor of $a_{k+1}$ and must be greater than $a_k$. Thus it can only be $a_{k+1}$ or $a_{k+1}/2$ (since $a_{k+1}/3=a_k$). If it is $a_{k+1}$, we're back to an integer ratio and will (possibly after some twos) encounter another integer ratio of at least $3$, taking the excess to at least $\frac{3}{2}\cdot\frac{3}{2}=\frac{9}{4}>2$. Thus the only case that could cause a problem is $\gcd(a_{k+2},a_{k+1})=a_{k+1}/2$.

Now consider the possible factors $r$ in $a_{k+2}=ra_{k+1}/2$. We must have $r\ge3$ since the sequence is strictly increasing. If $r=3$ or $r=4$, the ratio is $\frac{3}{2}$ or $2$ and thus ends the section without exhausting our accumulated excess of $\frac{3}{2}$, so this case is OK. If $r\ge6$, the ratio $a_{k+2}/a_{k+1}$ is at least $3$, giving us another excess factor of $\frac{3}{2}$ and taking us to $\frac{9}{4}>2$, so in that case we're also done. So the only case that could cause a problem is $r=5$, corresponding to a ratio of $\frac{5}{2}$. This gives us another excess factor of $\frac{5}{4}$.

But now the same arguments apply (with the ratio of $3$ replaced by $\frac{5}{2}$), so again we only have the options of going back to an integer ratio, ending the section with sufficient excess, taking the excess beyond $2$ or choosing $r=5$ for yet another ratio of $\frac{5}{2}$, which takes us back to the same situation. Eventually (in fact in just one more step), the excess factors of $\frac{5}{4}$ take the excess beyond $2$, and we're done.

Thus, every section necessarily accumulates an excess factor of $2$ that makes up for the low ratio at its end, so the inequality can never be violated.


问:
Number Theory : $11x \equiv44\pmod{64}$ how to solve?
how to solve this $11x\equiv 44\pmod{64}$ ?

attempt:

first of all need to find the inverst of $11x\equiv 44\pmod{64}$

by Euclid algorithm then:

$64=11\cdot 5 +9$

$11=1\cdot 9 +2$

$9=2\cdot4+1$

$1=9-2\cdot4$

what is the next step for the make the inverst I do not understand ?

答:
For fun, another approach you can take is Hensel lifting, which lets us find the roots of a polynomial $\mod 2$ and lift to a higher solution $\mod 2^n$. In order to do that, we need check that the criteria that you can find some $x$ such that $f(x)\equiv 0 \mod 2$ and $f'(x) \not \equiv 0 \mod 2$.

For that you define $f(x)=11x-44$ which we want the root(s) of. Then let's use $x=1$ as our guess, if that doesn't work we just try $x=0$.

$$ f(1) \equiv  11*1 - 44  \not \equiv 1 \mod 2$$
$$ f'(1) \equiv 11 \not \equiv 0 \mod 2$$

This fails the first criteria, since we require it to be $0 \mod 2$, so $1$ does not lift to a solution. Let's check $x=0$ now.

$$ f(0) \equiv  11*0 - 44  \equiv 0 \mod 2$$
$$ f'(0) \equiv 11 \not \equiv 0 \mod 2$$

We've succeeded in meeting the criteria for Hensel's lemma, and now we can apply the iteration formula which converges to the solution! We take as our guess $x_0 = 0$ and then iterate by,

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

So let's do that now,

$$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}= 0 - \frac{f(0)}{f'(0)} = - \frac{11*0-44}{11} = 4$$

Let's iterate again,

$$x_2=x_1-\frac{f(x_1)}{f'(x_1)} = 4 -\frac{f(4)}{f'(4)} = 4 -  \frac{11*4-44}{11} = 4$$

Since $x_2=x_1 = 4$ it means we've reached a fixed point and this is our final solution, we're done!

Some final remarks, in case you're wary about the $f'(x_n)$ evaluations in the denominator, the criteria we checked at the start end up satisfying the base case of an induction argument that makes up Hensel's lemma which ensures that we're only ever inverting a number that's relatively prime to $2$.

If you've never seen this before, you'll probably find it bizarre to see Newton's method used to solve modular arithmetic problems, and it turns out this is part of p-adic analysis. This gives you a way to get your hands on genuine p-adic numbers that satisfy some algebraic equation.


问:
If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that $\frac{1}{1-z}= \frac{1}{2}\left(1 + i\cot(\theta/2)\right)$

Question.(source) If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that
$$\frac{1}{1-z}= \frac{1}{2}\left( 1+ i\cot\frac{\theta}{2}\right)$$

I have been trying to rearrange the right-hand-side in order to make it look like the required form, but have been struggling to do so and want to see if anyone is able to show me what steps need to be taken to solve this question.

答:
Hint:

Rewrite the expression, using the exponential notation:
\begin{align}
\frac{1}{1-z}&= \frac{1}{1-\mathrm e^{i\theta}}= \frac{1-\mathrm e^{-i\theta}}{(1-\mathrm e^{i\theta})(1-\mathrm e^{-i\theta})}\\
&= \frac{1-\mathrm e^{-i\theta}}{(2-\mathrm e^{i\theta}-\mathrm e^{-i\theta})}
== \frac{1-\mathrm e^{-i\theta}}{2(1-\cos\theta)}\\
&=\frac{1-\cos\theta}{2(1-\cos\theta)}+i\,\frac{\sin\theta}{2(1-\cos\theta)}
\end{align}
continue with the duplication formulæ.


问:
Wrong solution in Hartshorne exercise III.2.7
I'm trying to solve Exercise III.2.7(a) in Hartshorne's Algebraic Geometry. And what I got is ${ H^1(S^1, \mathbb{Z}) = 0 }$ instead of expected ${ H^1(S^1, \mathbb{Z}) = \mathbb{Z} }$. My reasoning was the following:

$S^1$ is irreducible and $\mathbb{Z}$ is a constant sheaf, so by Exercise II.1.16(a), $\mathbb{Z}$ is flasque.

Applying Proposition III.2.5 we get ${ H^i(S^1, \mathbb{Z}) = 0 }$ for all ${ i > 0 }$. In particular, ${ H^1(S^1, \mathbb{Z}) = 0 }$.

Please help me find the mistake.

答: $S^1$ is given its usual topology in that exercise, which means it's not irreducible. For instance, $S^1\cap ([-1,0]\times[-1,1])$ and $S^1\cap ([0,1]\times[-1,1])$ are two proper closed subsets whose union is all of $S^1$.


问:
The euclidean space $\Bbb R^n$ is orientable as a manifold.
I know that

The euclidean space $\Bbb R^n$ is orientable as a manifold.

I think that it is orientable because it has a nowhere vanishing $n$-form.

But I am not sure.

Please can you explain me more formally and mathematically?

Thank you:)

答:
One very simple way of doing this is showing that

If a manifold has an atlas with exactly one chart, then it is orientable.

Of course, this applies to euclidean spaces.


问:
How to prove that $|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1})$ when $x,y \geq 0$ and $p \geq 1$?
For $x \geq 0$, $y \geq 0$, prove that

$$
|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1}).
$$

I thought it would be simple but I messed everything up. Here are my attempts.

Fixing $x > y \geq 0$, I thought about the function

$$
h(p)=x^p-y^p-p(x-y)(x^{p-1}+y^{p-1})
$$

Then $h(1)=0$ and I wanted to prove that $h(p) \leq 0$ when $p \geq 1$ by discussing its derivative but the derivative is messy offering no way out.

Also I thought about after fixing $y \geq 0$, consider the function

$$
\varphi(x)=x^p-y^p-p(x-y)(x^{p-1}+y^{p-1})
$$

where $\varphi(y)=0$ and I want to prove that $\varphi(x) \leq 0$ when $x \geq y$ yet again it's a messy way.

I also thought about restricting $p$ to $\mathbb{N}$ or $\mathbb{Q}$ using decomposition like

$$
(x^p-y^p)=(x-y)(x^{p-1}+x^{p-1}y+x^{p-2}y^2+\cdots+y^{p-1})
$$

but the number of terms does not match.

But I do believe this should be a easy question with some special background. I must have missed something critical. Any hint/solution appreciated!

答:
It's an direct consequence of the Hermite-Hadamard inequality

We have with $f(x)=px^{p-1}$ a convex function , $x\geq y\geq 0$ and $p\geq 1$:

$$\frac{1}{x-y}\int_{y}^{x}f(x)dx\leq \frac{f(x)+f(y)}{2}$$

Or :

$$\frac{x^p-y^p}{x-y}\leq \frac{px^{p-1}+py^{p-1}}{2}$$


问:
Solve $\int\left(\ln5x+2e^{2x}\right)\,dx$
$$\int\left(\ln(5x)+2e^{2x}\right)\,dx$$

I am not getting how to integrate this function. Please help.

答:
You have
$$\int (\ln 5x + e^{2x})~dx = \int \ln 5x~dx + \int 2e^{2x}~dx = I_1 + I_2.$$
For $I_2$ try by subtstitution and for $I_1$ try by parts with $u = \ln 5x$ and $v' = 1$.


答: Doing it by parts and splitting, We get $\int\left(\ln(5x)+2e^{2x}\right)\,dx=\boxed{x\ln \left(5x\right)-x+e^{2x}+C}$

Title: Solve $\int\left(\ln5x+2e^{2x}\right)\,dx$

Body:
$$\int\left(\ln(5x)+2e^{2x}\right)\,dx$$

I am not getting how to integrate this function. Please help.


答:
Here is a fun trick I learned when I saw this question for the first time.

$$\frac{1}{1-e^{i\theta}} = \frac{e^{-i\theta/2}}{e^{-i\theta/2} - e^{i\theta/2}} = \frac{\cos\tfrac12\theta - \sin\tfrac12\theta}{-2i\sin\tfrac12\theta} = \ldots$$

Title: If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that $\frac{1}{1-z}= \frac{1}{2}\left(1 + i\cot(\theta/2)\right)$

Body:
Question.(source) If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that
$$\frac{1}{1-z}= \frac{1}{2}\left( 1+ i\cot\frac{\theta}{2}\right)$$

I have been trying to rearrange the right-hand-side in order to make it look like the required form, but have been struggling to do so and want to see if anyone is able to show me what steps need to be taken to solve this question.


问:
How to calculate the number of subgroups of a group whose order is $19$?
How to calculate the number of subgroups of a group whose order is $19$?

Normally, I'd resolve this problem with the Lagrange's Theorem, but I don't know how to do it with this one because I only know the order.

So this is not the best approach.

What's the answer to this problem?

Thanks.

答: If the main group has order $19$, which is a prime, then it can only be the cyclic $\mathbb Z_{19}$. Its only subgroups are the trivial group and itself.


答: I think your approach with Lagrange's Theorem is a correct one. Let $G$ be a group such that $|G| = 19$ and $H \le G$. Then, By Lagrange's Theorem, we have $|H|\  \big |\ |G| = 19$. But $19$ is a prime number so only divisors of $19$ are $1$ and $19$. If $|H| = 1$, then clearly $H$ is trivial group. If $|H| = 19$, then $H = G$.

Body:
How to calculate the number of subgroups of a group whose order is $19$?

Normally, I'd resolve this problem with the Lagrange's Theorem, but I don't know how to do it with this one because I only know the order.

So this is not the best approach.

What's the answer to this problem?

Thanks.


问:
Probablility of a dart landing closer to the center than the edge of a square dartboard?
The question is: There is a square dart board with a side length of $2$m, and a dart has equal probability to land anywhere on the board. What is the probability that the dart will land closer to the center than to the edges? So I know that if I center the square at the origin, I have to find $f(x)$ such that every point on $f(x)$ is equidistant between the origin and the line $x=1$ or $y=1$ (I'm only working with the first quadrant since by symmetry the probability should be the same). Another reasonable assumption that I made is that $f(x)$ is reflected over $y=x$, so I only have to work with the first $45$ degrees of the first quadrant. So to find $f(x)$ I did the following: $\sqrt{x^2+[f(x)]^2}=1-x$, so $f(x)=\sqrt{1-2x}$. Now I know that I just find the area under this curve and divide it by $1/8$ the area of the square to find the probability. I just wanted to know if my $f(x)$ was okay? Thanks! </p>

答:
Your thought process looks good, but the integration is actually a bit tricky.  Here is your dartboard:

We'd like to calculate the area of the center shape, and as you say, we can split this into eight slices of equal area.  Here we see the two slices in the first quadrant:

Let's compute the area of the slice from $45^{\circ}$ to $90^{\circ}$:

$$\int_{0}^{\sqrt{2}-1}(\frac{1-x^2}{2}-x)dx=\frac{1}{6}(4\sqrt{2}-5)$$

Multiplying this by $8$ gives us the area of the center shape.  Then, we divide by $4$, the area of the dartboard.  The final probability is then:

$$\frac{1}{3}(4\sqrt{2}-5)\approx.2190$$

Judging by our picture above, this seems right.  The center shape is about $1/5$ the area of the dartboard.


答:
Let $H\leq G$ be a subgroup of $G$. From Lagrange's Theorem $|H|\mid |G|$. Since $|G|=19$ we have that either $|H|=1$ or $|H|=19$. Let's consider both cases:

$|H|=1$. In this case the subgroup $H$ has only one element. Every group has to have an identity element, so $H=\langle e\rangle$.

$|H|=19$. Let's see what this group looks like. Let $g\in H$, the order of $g$, $|g|$, divides $|H|$, so $|g|\mid 19$. Only on element has order $1$, which is the identity. If $g\neq e$, then $|g|=19$ which means that $g$ is a generator, $H=\langle g\rangle$. Thus we can make an isomorphism $$\phi: H\rightarrow\Bbb Z_{19}\\g\mapsto 1$$

Title: How to calculate the number of subgroups of a group whose order is $19$?


问:
Characterizing units in polynomial rings
I am trying to prove a result, for which I have got one part, but I am not able to get the converse part.

Theorem. Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\dots,a_{n}$ are all nilpotent in $R$.

Proof. Suppose $f(X)=a_{0}+a_{1}X+\cdots +a_{n}X^{n}$ is such that $a_{0}$ is a unit in $R$ and $a_{1},a_{2}, \dots,a_{r}$ are all nilpotent in $R$. Since $R$ is commutative, we get that $a_{1}X,a_{2}X^{2},\cdots,a_{n}X^{n}$ are all nilpotent and hence also their sum is nilpotent. Let $z = \sum a_{i}X^{i}$ then $a_{0}^{-1}z$ is nilpotent and so $1+a_{0}^{-1}z$ is a unit. Thus $f(X)=a_{0}+z=a_{0} \cdot (1+a_{0}^{-1}z)$ is a unit since product of two units in $R[X]$ is a unit.

I have not been able to get the converse part and would like to see the proof for the converse part.

答:
Let $f=\sum_{k=0}^n a_kX^k$ and $g= \sum_{k=0}^m b_kX^k$. If $f g=1$, then clearly $a_0,b_0$ are units and:

$$a_nb_m=0 \tag1$$

$$a_{n-1}b_m+a_nb_{m-1}=0$$

(on multiplying both sides by $a_n$)

$$\Rightarrow (a_n)^2b_{m-1}=0 \tag2$$

$$a_{n-2}b_m+a_{n-1}b_{m-1}+a_nb_{m-2}=0$$
(on multiplying both sides by $(a_n)^2$)
$$\Rightarrow (a_n)^3b_{m-2}=0 \tag3$$
$$.....$$
$$.....+a_{n-2}b_2+a_{n-1}b_1+a_nb_0=0$$
(on multiplying both sides by $(a_n)^m$)
$$\Rightarrow (a_n)^{m+1}b_{0}=0 \tag{m+1}$$

Since $b_0$ is an unit, it follows that $(a_n)^{m+1}=0$.

Hence, we proved that $a_n$ is nilpotent, but this is enough. Indeed, since $f$ is invertible, $a_nx^n$ being nilpotent implies that $f-a_nX^n$ is unit and we can repeat (or more rigorously, perform induction on $\deg(f)$).

OwnerDisplayName: anonymous


问:
Solid spaces are locally contractible
I have a question while reading Steenrod's The Topology of Fibre Bundles, section 12.

A space $Y$ is called solid if, for any normal space $X$, closed subset $A$ of $X$, and map $f:A\to Y$, there exists a map $f':X\to Y$ such that $f'|_A=f$.

Let $Y$ be solid such that $Y\times I$ is normal. Fix a point $y_0\in Y$. Note that $A:=(Y\times 0)\cup (y_0\times I)\cup (Y\times I)$ is a closed subset of $Y\times I$. Define $f:A\to Y$ by $f(y,0)=y$, $f(y,1)=y_0$ and $f(y_0,t)=y_0$. Then solidity of $Y$ implies that $f$ extends to $f':Y\times I\to Y$. Now $f'$ is a homotopy from $\textrm{id}_Y$ to the constant map $Y\to y_0$. Thus $Y$ is contractible. Since $y_0$ is arbitrary, it also follows that $Y$ is locally contractible.

I can't see why $Y$ is locally contractible. How does this argument shows that each point of $Y$ have arbitrary small locally contractible neighborhoods?

答:
A more common notation for a solid space is "absolute extensor for normal spaces".

Your construction of $f'$ shows that $(Y,y_0)$ is pointed contractible for each $y_0 \in Y$. This immediately implies that

For each open neigborhood $U$ of $y_0$ in $Y$ there exists an open neighborhood $V$ of $y_0$ in $Y$ contained in $U%
$ such that the inclusion $V \hookrightarrow U$ is null-homotopic.

If this property is satisfied, $Y$ is called locally contractible at $y_0$. If $Y$ is locally contractible at all of its points, it is called locally contractible.

This is the standard definition. The requirement that each $y_0 \in Y$ has arbitrarily small (open) contractible neigborhoods is stronger and I doubt that is true for all absolute extensors. You should check Steenrod's definition.

See also ANR is locally contractible .

Title: Solid spaces are locally contractible


问:
Showing subset is dense
Let $X$ be finite discrete metric space. Since it is finite and discrete every map from $X$ to real numbers will be bounded and continous. Let $C$ be the set of such functions. I want to show that $C$ is seperable where $C$ is metric space with respect to the norm $||f|| = \sup{f}$.

I consider the set $B =\{f :  \text{the image of $f$ is the subset of rational numbers}\}$

I feel that $C$ has same cardinality as $\mathbb{R^n}$ and $B$ has same cardinality as $\mathbb{Q^n}$ where $n$ is cardinality of $X$. Since $\mathbb{Q^n}$ is seperable in $\mathbb{R^n}$, is it enough to prove that these sets have same cardinal numbers as mentioned above and conclude that $B$ is seperable in $C$?

Or

How can I show that any function which is not in $B$ is limit point of $B$?

答:
Regarding your comment : Consider $X$ to be the Sorgenfrey plane and $Y$ to be its anti-diagonal. Then, even though $X$ and $Y$ have the same cardinality and $X$ is separable, the same does not hold for $Y$.

Regarding your question : I read $B=\lbrace f\in C : f(X)\subseteq\mathbb{Q}\rbrace$. Let $f\in C$ and $\epsilon >0$. Pick $q_i\in\mathbb{Q}$ such that $|f(x_i)-q_i|<\frac{\epsilon}{2}$ for every $x_i\in X=\lbrace x_1,... x_n\rbrace$. Then, $g\in B$ such that $g(x_i)=q_i$ is such that $||g-f||<\epsilon$

Title: Showing subset is dense


问:
Commutator subgroup of $S_n$ generated by commutators of transpositions
Is it true that the commutator group $[S_n, S_n]$ is generated by elements of the form $[\tau_1, \tau_2]$ for $\tau_1, \tau_2$ transpositions?

I know that in general, when a group $G$ is generated by a set $S$, then one has to take the normal closure of the group generated by commutators of elements in $S$ to obtain $[G,G]$.

答:
The commutator of $(1,3)$ and $(2,3)$ is $(1,2)(2,3)=(1,2,3)$, so any $3$ cycle is a commutator of two transpositions.
For $S_3$ we are done.

Now for $n\geq 4$, it is know that every prouct of an even number of transposition (an alternate permutation) is a product of 3 cycle.

Indeed if the two transpositions have disjoint support $(12)(34)= (123)(234)$, and if not $(12)(23)=(123)$

Now it is known that the derived group of $S_n, n\geq 4$  (the alternate group) is generated by the 3-cycles.

So any product of an even number of transposition is a product of 3 cycles, and a product of commutator of transposition. But the derived group of $S_n$ is the kernel of the signature, so the set of permutation which are the product of an even number of transposition..


问:
Prove the direct sum: $V = im(T)⊕im(S)$
Let there be V vector space.
we are given two linear transformations - $S,T : V \to V$ such that:

$T + S = Iv$

$T \circ S = 0 $

$T \circ T  = 0 $

Using all of the above - I need to prove that:

$V = im(T)⊕im(S)$

I have tried to use the Rank–nullity theorem with no success, any help?

答:
It is clear that $im(T)+im(S)\subseteq V$. To show that $V \subseteq im(T)+im(S)$, pick $v \in V$. Now, by (1), $T(v)+S(v)=I(v)=v$. So, $v \in im(T)+im(S)$ and hence $V \subseteq im(T)+im(S)$.

To show that the sum is direct, pick $v \in im(T) \cap im(S)$. Then, by definition, there exists $a \in V$ such that $T(a)=v$. Now, by (1), $T(a)+S(a)=I(a)=a$ and so, by (2) and (3), $0=0+0=T(T(a))+T(S(a))=T(T(a)+S(a))=T(a)=v$, thus, $v=0$ and we are done.

Title: Prove the direct sum: $V = im(T)⊕im(S)$


问:
On the convergence of improper integrals
Question

Determine the value of $r$ for which the integral $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is convergent and evaluate the integral for that value of $r$.

My working

It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0\ $.

If $r \neq 0\ $, then

\begin{align}
\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & =
\lim\limits_{a\to\infty}[r\ln(x + 1) - \frac 3 4 \ln|2x^2 + r|]^{x = a}_{x = 1}
\\[5 mm] & =
\lim\limits_{a\to\infty}[r\ln(a + 1) - \frac 3 4 \ln|2a^2 + r| - r\ln 2 + \frac 3 4 \ln|2 + r|]
\\[5 mm] & =
\lim\limits_{a\to\infty}\left(r\ln\frac {a + 1} 2 + \frac 3 4 \ln\left|\frac {2 + r} {2a^2 + r}\right|\right)
\end{align}

This question just appeared in my final calculus examination this afternoon and I got stuck here ):. We have practiced such questions before but I cannot seem to get this one. I would usually proceed by trying L'Hopital's Rule since we want the limit to exist. I know I probably will not be touching calculus for quite a while from now, but I still want to figure this out. Any intuitions as to how I should proceed will be greatly appreciated :)

答:
Try to bring the logarithms together! :)

$$\lim_{a\to \infty} \biggl(r\ln\frac{a+1}{2}+\frac 3 4\ln|\frac{2+r}{2a^2+r}|\biggl)$$
You maybe should take $r < -2$ and $r > -2$ because of the absolute value, but it gives the same result so I'll go with normal brackets.
$$\lim_{a\to \infty} (\ln(\frac{a+1}{2})^r+ \ln(\frac{2+r}{2a^2+r})^\frac 3 4 )$$
$$\lim_{a\to \infty} \biggl(\ln\bigl((\frac{a+1}{2})^r \cdot (\frac{2+r}{2a^2+r})^\frac 3 4\bigl )\biggl)$$

$$\lim_{a\to \infty} \biggl(\ln\bigl((\frac{(2+r)^\frac 3 4}{2^r}) \cdot (\frac{(a+1)^r}{(2a^2+r)^\frac 3 4})\bigl )\biggl)$$
Now you can split the logarithms if you want, take the minus with the left part when you do $r < -2$. The first part is a constant. The second part is the intertesting one. The limit of the logarithm of that part has to exist. Both numerator and denominator are polynomials of $a$.

If the degree of the top polynomial ist bigger, the fraction will go to infinity and so does the logarithm of it. The integral diverges.

If the degree of the bottom polynomial is bigger, the fraction will go to $0$ and the logarithm to negative infinity. The integral diverges.

Only if both polynomials have the same degree, the fraction will go to a finite number and so does the logarithm. The integral converges. Comparing the degrees gives:
$$r = 2*3/4$$
So $r=1.5$.
Getting the value of the integral being $0.75\ln(7)-3\ln(2)$ should not be a problem.

Its my first answer on math.stackexchange so my formulas look pretty bad and maybe my answer isnt perfect but I hope I could help you! :P


问:
Integrate: $\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)}dx$
How to evaluate the following
$$\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)} dx $$
Given hints says to construct a rectangle $0\to R\to R+i\to i \to 0$ and consider $\displaystyle f(z):=\frac{e^{iaz}}{e^{2\pi z}-1} $ and evaluate around it but that does not help.

ADDED:: I need to evaluate it with method of contour.

Particularly using given hint on book, when I integrate $i \to 0$, the integral does not converge. Also, real part of $\displaystyle f(z):=\frac{e^{iaz}}{e^{2\pi z}-1} $ does not converge around $0$.

Also I considered $\displaystyle f(z):=\frac{e^{iaz}-e^{-iaz}}{e^{2\pi x}-1} $, and integrate from $-R \to R \to R+i \to -R+ i \to R$ and I get $\displaystyle \int_{-\infty}^\infty f(z) dz$ but the function is not symmetric about $0$.

If I consider $\displaystyle f(z):=\frac{e^{iaz}}{e^{2\pi x}-1} $ around the path $-R - i \to R - i \to R + i \to -R +i \to -R -i $, I again end up with $\displaystyle \int_{-\infty}^{\infty}f(z) dz$ and I cannot extract $\int_0^{\infty}$ due to  function is not symmetric at $0$ because of $e^{2\pi z}$ at the denominator.

Please help!!

答:
Well, it's not the direct answer on you questions (because you were given a direct hint in the link above), but an alternative way.
You can look at it like the Laplace transform:
$$\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)} dx=2\int_0^{\infty} \frac{\sin (ax)}{1-e^{-2\pi x}}e^{-2\pi x} dx=2\int_0^{\infty} f(x)e^{-sx} dx$$
with $f(x)=\frac{\sin (ax)}{1-e^{-2\pi x}}$ and $s=2\pi$.
The Laplace transform for $f(x)$ (valid when $\left| \Im(a)\right| <2 \pi$):
$$ \mathcal{L}\left\{\frac{\sin (ax)}{1-e^{-2\pi x}}\right\}=\frac{\psi ^{(0)}\left(\frac{i a+s}{2 \pi }\right)-\psi ^{(0)}\left(\frac{-i a+s}{2 \pi }\right)}{4 \pi  i}$$
where $\psi ^{(0)}(\cdot)$ is the digamma function.
Setting $s=2\pi$, multiplying by 2 and simplifying one can obtain:
$$\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)} dx=\frac{1}{2}\coth\left(\frac{a}{2}\right) -\frac{1}{a}$$</p>


问:
Why is the Gram-Schmidt process used?
Why is this process used to construct an orthonormal basis? Why can’t an orthonormal basis be found from a basis by just taking the norm of each of each and dividing the respective basis vector by it?The notes I have on this just state it is used, and shows the method</p>

答: If you were to just divide each vector in a basis by their norm, they would be unit norm, but not orthogonal.</p>

Title: Why is the Gram-Schmidt process used?


问:
How to prove there is only one positive root?
If I want to prove $x^n+x^{n-1}-a=0$ has only one positive root for $a> 0$ and $n \ge 2$
Can I say for $n=2$ : $x^2+x-a=0$ and we know only one of the roots is positive.
Now if we we know there is only one root for $x^n+x^{n-1}-a=0$
now I prove for $x^{n+1}+x^{n}-a=0$
In the end I get to $x=1$ , but I don't think I’m doing it correctly.
Should I use different $a$ s?

Is my way wrong? If it is, could you tell me what I should do?

答:
Let $p(x) = x^n + x^{n-1} - a.$

That it has at least one root is clear by Intermediate Value Property. (The value at $0$ is negative and $p(x) \to \infty$ as $x \to \infty$.)

Now, note that $p'(x) = nx^{n-1} + (n - 1)x^{n-2} > 0$ for all $x > 0.$

Thus, if $p$ had two positive roots $a < b$, then $p'(c) = 0$ for some $a < c < b$, by Rolle's theorem. But $c > 0$ and hence, $p'(c) = 0$ is not possible.


问:
Trying to determine the determinant of an abstract matrix
I'm trying to write about linear homogeneous recurrence relations and I've come up on the following matrix :$$A=\begin{pmatrix}
c_1 & c_2 & \cdots & c_{k-1} & c_k\\
1 & 0 & \cdots & 0 & 0\\
0 & 1 & \ddots & 0 & 0\\
\vdots & \vdots & \ddots & \ddots & \vdots\\
0 & 0 & \cdots & 1 & 0
\end{pmatrix},$$ where $c_1,\cdots c_k$ is a real number. I need to find the eigenvalues of this matrix. So far I've got that $$c_A(r)=\begin{vmatrix}
c_1-t & c_2 & \cdots & c_{k-1} & c_k\\
1 & -t & \cdots & 0 & 0\\
0 & 1 & \ddots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1 & -t
\end{vmatrix},$$ I need to express it in its polynomial form but I can't. I've done it for $k=3$ and found $c_A=-r^3+c_1r^2+c_2r +c_3.$ I want to show that $c_A(r)= r^k - \sum_{i=1}^{k-1}c_{i}r^{k-i}$.
Any help ?

答:
Such a matrix (or rather, once we reversed the order of rows and columns, and possibly taken the transpose) is called a companion matrix.

The easiest way to compute the determinant is to expand along the last column. This gives $(-t)$ times the determinant of a smaller matrix of the same type, plus $(-1)^nc_k$. The result now follows by induction on $k$.


问:
Ratio in triangle $\Delta ABC$. Find ratio $|CF|:|FP|$.
Let $\Delta ABC$ be triangle and  $P$ and $Q$ points on $AB$ and $AC$ such that $|AP|:|PB|=1:5$ and $|AQ|:|QC|=2:1$. Lines $BQ$ and $CP$ intersect in point $F$. Find ratio $|CF|:|FP|$.

I want to solve this problem without using triangle similarities or some theorems, but only vectors. I tried to write vector $\vec{CF}$ and $\vec{CP}$ using for example $\vec{AB}$ and $\vec{AC}$ that would solve my problem, but can't do it.

答:
There is a one way...
let's hang 1 kg item on A after that you can easily say that u must hang 2 kg item on C to  balance AC line where Q is center of mass. ( mass(A) * AQ = QC * mass(c) )
using the same way  find mass of P
mass(A) * AP = PB mass(B)  $ \rightarrow $ mass(B) = 1\5 so the center of mass$ (P) = (1 + 1/5) $

Now you know the mass of P and C
now write mass(P)*PF = mass(C) * CF
$ \rightarrow $  $ \frac{CF}{FP} = \frac{mass(P)}{(mass(C)} $


答:
I have finally managed to answer the question :)

It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0$.

If $r \neq 0$, note that we can always choose $a$ large enough such that $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x = \int^a_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x + \int^{\infty}_a (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x,$$ where $$\int^a_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is always convergent $\forall\ r \in \mathbb {R}$.

Then, the problem reduces to checking the convergence of $$\int^{\infty}_a (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x,$$ where

\begin{align}
\int^{\infty}_a (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & =
\lim\limits_{b\to\infty}[r\ln(x + 1) - \frac 3 4 \ln(2x^2 + r)]^{x = b}_{x = a}
\\[5 mm] & =
\lim\limits_{b\to\infty}[r\ln(b + 1) - \frac 3 4 \ln(2b^2 + r) - r\ln (a + 1) + \frac 3 4 \ln(2a^2 + r)]
\\[5 mm] & =
\ln \left[\lim\limits_{b\to\infty}\frac {(b + 1)^r} {(2b^2 + r)^{\frac 3 4}}\right] - r\ln (a + 1) + \frac 3 4 \ln(2a^2 + r)
\\[5 mm] & =
\ln \left[\lim\limits_{b\to\infty}\frac {b^{r - \frac 3 2}(1 + \frac 1 b)^r} {(2 + \frac r {b^2})^{\frac 3 4}}\right] - r\ln (a + 1) + \frac 3 4 \ln(2a^2 + r).
\end{align}

Observe that, due to the condition imposed on $a$, the last two terms, namely, $- r\ln (a + 1)$ and $\frac 3 4 \ln (2a^2 + r)$ always exist. Moreover, as $b \rightarrow \infty$, $(1 + \frac 1 b)^r$ and $(2 + \frac r {b^2})^{\frac 3 4}$ are always convergent, so it suffices to check the convergence of $\lim\limits_{b\to\infty}b^{r - \frac 3 2}$ and it is easy to see that $r$ can only take on one value, that is $\frac 3 2$.

Thus, when $r = \frac 3 2$,

\begin{align}
\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & =
\lim\limits_{c\to\infty}[\frac 3 2\ln(x + 1) - \frac 3 4 \ln(2x^2 + \frac 3 2)]^{x = c}_{x = 1}
\\[5 mm] & =
\lim\limits_{c\to\infty}[\frac 3 2\ln(c + 1) - \frac 3 4 \ln\frac {4c^2 + 3} 2 - \frac 3 2\ln 2 + \frac 3 4 \ln\frac 7 2]
\\[5 mm] & =
\frac 3 4\lim\limits_{c\to\infty}[\ln(c + 1)^2 - \ln(4c^2 + 3)] - \frac 3 2\ln 2 + \frac 3 4 \ln 7
\\[5 mm] & =
\frac 3 4\lim\limits_{c\to\infty}\ln\frac {(c + 1)^2} {4c^2 + 3} - \frac 3 2\ln 2 + \frac 3 4 \ln 7
\\[5 mm] & =
\frac 3 4\ln\frac 1 4 - \frac 3 2\ln 2 + \frac 3 4 \ln 7
\\[5 mm] & =
\frac 3 4 \ln 7 - 3\ln 2
\end{align}

Title: On the convergence of improper integrals

Body:
Question

Determine the value of $r$ for which the integral $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is convergent and evaluate the integral for that value of $r$.

My working

It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0\ $.

If $r \neq 0\ $, then

\begin{align}
\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & =
\lim\limits_{a\to\infty}[r\ln(x + 1) - \frac 3 4 \ln|2x^2 + r|]^{x = a}_{x = 1}
\\[5 mm] & =
\lim\limits_{a\to\infty}[r\ln(a + 1) - \frac 3 4 \ln|2a^2 + r| - r\ln 2 + \frac 3 4 \ln|2 + r|]
\\[5 mm] & =
\lim\limits_{a\to\infty}\left(r\ln\frac {a + 1} 2 + \frac 3 4 \ln\left|\frac {2 + r} {2a^2 + r}\right|\right)
\end{align}

This question just appeared in my final calculus examination this afternoon and I got stuck here ):. We have practiced such questions before but I cannot seem to get this one. I would usually proceed by trying L'Hopital's Rule since we want the limit to exist. I know I probably will not be touching calculus for quite a while from now, but I still want to figure this out. Any intuitions as to how I should proceed will be greatly appreciated :)


问:
Properties of basic set theory
The question is about a set:
$$B=\{a_1,a_2,a_3,...,a_n\} \subset \mathbb R$$
And would like to know how to calculate $B^n$ where $n \in\Bbb N$?</p>

答:
For a set $S$, we often write $S^n$ to denote the $n$-fold cartesian product of $S$ with itself. In your case, we would have
$$
B^n = \{(b_1, b_2,\ldots, b_n)\mid b_i = a_{i_k}, a_{i_k}\in B\}.
$$
In other words, $B^n$ consists of ordered $n$-tuples, where each element in the tuple is an element of the original $B$.</p>

Title: Properties of basic set theory


问:
How to draw a $(f \circ g)(x)$ function without a rule and with graphs of $f(x)$ and $g(x)$?
Is it possible to draw the composite function $(f \circ g)(x)$ without having a rule for the two functions $f$ and $g$ and only having graphs for these two functions?

For example these two functions

答:
It is possible but a bit tedious. See below the gif showing how you may construct the composition $f\circ g$ point by point using the line $y=x$.

I chose the functions to look similar to those in your picture:


问:
Finite Concepts that Require Infinity.
Answers to this question, I hope, would address the objections one might have towards the (many different) foundations of infinity by examples of finite concepts - theorems, definitions, intuitions, perspectives, etc. - that require the infinite, whether that be in a proof, an important source of motivation, a key example (like some group like $S_3$ can be for non-abelian group theory), etc.

The examples-counterexamples tag could preempt criticism that the question is too broad.

What objections?

Well, for example, there're those that posit that infinity has no place in mathematics as, otherwise, we would be - and I quote - "quantifying the unquantifiable".

The Question:

What examples are there of finite concepts where infinity does some essential heavy-lifting?

Thoughts:

Nothing springs to mind.

An example might take the following form:

Theorem: (Only finite concepts hypothesised.) Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. (Finite consequences.)

Proof: Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. (Infinity.) Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum. $\square$

(I used filler text but I hope this captures the sort of thing I'm looking for, without belabouring the point.)

What I want is a bunch of things to point to and say, "here: important finite things can require infinity" - and bring something to the table that is tangible in an area that is largely seen as intangible.

A Potential Example:

Finitely-presented groups can be infinite. For example, the presentation $$\langle a\mid \rangle$$ is finite, whereas the group $(\Bbb Z, +)$ it corresponds to is infinite.

Preliminary Discussion:

Before asking here, I brought it up in chat.

Please help :)

答:
Goodstein's Theorem qualifies as a theorem about the integers that can't be proved without using transfinite ordinals. Rather than attempt to summarize it here, I'll point to the Wikipedia essay linked in my comment on the original post, and also to some earlier discussions here on math.stackexchange:

Goodstein sequences

Do we know if there exist true mathematical statements that can not be proven?

Goodstein's sequences and theorem.

Counterintuitive Goodstein's Theorem

Goodstein's theorem

and a LOT of others.

Title: Finite Concepts that Require Infinity.


问:
Test and probability / Bayes
There is a virus test with has 95% reliability. We randomly choose one person from a country with 10,000,000 people, in which it is estimated that the number of cases of this virus is 20,000.
The selected person is tested positive. What is the probability that this person is actually positive?

This is clearly an example of Bayes theorem use, but I am getting an odd result:

$\mathrm{P}(A \mid Β) = \frac {\mathrm{P}(B \mid A).\mathrm{P}(A)}{\mathrm{P}(B)}$

$\mathrm{P}(A) = \frac {20,000}{10,000,000} = 0.002$

Since the test has 95% reliability, in the entire country (with population 10,000,000), if all were tested, there would be 95,000 people who are sick test positive and 95,000 of the healthy people test negative, therefore a total of 190,000. So:

$\mathrm{P}(B) = \frac {190,000}{10,000,000} = 0.019$ (unconditional probability).

Finally, since the sensitivity of the test is 95%, $\mathrm{P}(B \mid A) = 0.95$.

So from Bayes theorem we get

$\mathrm{P}(A \mid Β) = \frac {0.95*0.002}{0.019} = 0.1$, that is, 10%.

But this is a bit counter-intuitive.

Is it correct? Thank you

答:
There is a virus test with has 95% reliability.

This sentence is ambiguous. Reliability of a test is a mixture of Sensibility and Specificity.

Let's assume that you mean Sensitivity and Specificity both $95\%$

Thus the number of  positive tests is

$$0.95\times 20,000+0.05\times 9,980,000=19,000+499,000=518,000$$

But only $19,000$ are actually positive, thus the PPV (Positive Predictive Value) is

$$PPV=\frac{19,000}{518,000}\approx 3.67\%$$

The result is not counterintuitive because the correct predictive value depends not only on the test reliability but also on the Virus prevalence in the population. That is the reason why mass test on all the population are not recommended.

Try the same exercise with a prevalence of 40% and realize that your PPV is about 93%.

For  your exercise, the PPV obviously increases if you repeat the test and it is always positive

$$PPV(1)\approx 3.67\%$$

$$PPV(2)\approx 41.98\%$$

$$PPV(3)\approx 93.22\%$$

$$PPV(4)\approx 99.62\%$$

Where with $PPV(i)$ I mean the probability to have the virus after $i$ consecutive positive tests

Title: Test and probability / Bayes


问:
ANR is locally contractible
Recall that a space $X$ is contractible if there exists a homotopy $h:X\times [0,1]\to X$ such that $h$ is equal to the identity map on $X\times\{0\}$ and $h$ is constant on $X\times\{1\}$.

Please help me out since I'm stuck with this question, namely, why each ANR is locally contractible. It seems that I can prove that each AR is contractible this way: Consider $A = X\times \{0\} \cup X\times \{1\} \cup \{x_0\}\times [0,1] \subset X\times [0,1]$, where $x_0$ is any point from $X$. This set is closed in $X\times [0,1]$. Define $h: A\to X$ by $h|_{X\times \{0\}} = id_X$, $h|_{X\times \{1\}} = x_0$, $h|_{\{x_0\}\times [0,1]} = x_0$. This function is continuous, we can extend it (since being an AR is equivalent to being an AE) to the whole space $X\times [0,1]$. And this will be the required homotopy that squeezes $X$ into a point ($x_0$).

But similar approach seems not to work in case of an ANR. Because here I get a neighborhood $V$ of $A$ in $X\times [0,1]$ and an extension of $h$ onto this neighborhood. I can of course find an open $U\subset X$ such that $x_0\times [0,1] \subset U\times [0,1]\subset V$ (where $x_0$ is the point for which I want to find a contractible neighborhood) and consider the restriction of my extension onto $U\times [0,1]$, but the problem is that this map is not necessarily a map from $U\times [0,1]$ to $U$ (the range of my restricted extension can be larger than $U$).

Can anybody please give a hint on what's going on? (a hint would be even better than a full answer).

答:
It is surprising that your question has been unanswered for six years.

It seems that there are two rivalling definitions of "locally contractible":

$Y$ is locally contractible if each $y_0 \in Y$ has arbitrarily small (open) contractible neigborhoods.

For each $y_0 \in Y$ and each open neigborhood $U$ of $y_0$ in $Y$ there exists an open neighborhood $V$ of $y_0$ in $Y$ which is contained in $U$ such that the inclusion $V \hookrightarrow U$ is null-homotopic.

In my opinion 2. is the standard definition. Clearly 1. implies 2., but I doubt that the converse is true. You have proved that ANRs are locally contractible in the sense of 2.

Many concrete examples of locally contractible spaces satisfy 1., e.g. manifolds and CW-complexes.

The concept of local contractibility was introduced by K. Borsuk in the nineteen-thirties. See

Borsuk, K. "Über eine Klasse von lokal zusammenhängenden Räumen." Fund. Math 19 (1932): 220-242.

Also see

Borsuk, Karol. Theory of retracts. Vol. 44. Państwowe Wydawn. Naukowe, 1967.

Title: ANR is locally contractible


答:
$20000$ are sick of the virus and $9,980,000$ are not sick.

Test has $95\%$ reliability.

Out of $20,000$ sick people -

$19,000$ will test positive and $1000$ will test negative.

Out of $9,980,000$ healthy people,

$499,000$ will test positive and $9,481,000$ will test negative.

Unconditional probability of being sick $P(A) = 0.002$

Unconditional probability of testing positive $P(B) = \frac{499,000+19,000}{10,000,000} = 0.0518$

Probability of testing positive given the person is sick $P(B|A) = 0.95$

So probability of a person really being sick if tested positive,

$\, P(A|B) = \frac{0.95 \times 0.002}{0.0518} \approx 3.67\%$

Title: Test and probability / Bayes

Body:
There is a virus test with has 95% reliability. We randomly choose one person from a country with 10,000,000 people, in which it is estimated that the number of cases of this virus is 20,000.
The selected person is tested positive. What is the probability that this person is actually positive?

This is clearly an example of Bayes theorem use, but I am getting an odd result:

$\mathrm{P}(A \mid Β) = \frac {\mathrm{P}(B \mid A).\mathrm{P}(A)}{\mathrm{P}(B)}$

$\mathrm{P}(A) = \frac {20,000}{10,000,000} = 0.002$

Since the test has 95% reliability, in the entire country (with population 10,000,000), if all were tested, there would be 95,000 people who are sick test positive and 95,000 of the healthy people test negative, therefore a total of 190,000. So:

$\mathrm{P}(B) = \frac {190,000}{10,000,000} = 0.019$ (unconditional probability).

Finally, since the sensitivity of the test is 95%, $\mathrm{P}(B \mid A) = 0.95$.

So from Bayes theorem we get

$\mathrm{P}(A \mid Β) = \frac {0.95*0.002}{0.019} = 0.1$, that is, 10%.

But this is a bit counter-intuitive.

Is it correct? Thank you


问:
Cross section of cylinder $5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$
Prove that

$$5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$$

represents a cylinder whose cross-section is an ellipse of eccentricity $\frac{1}{\sqrt 2}$.

I know how to find the plane of a cross section if I know the direction ratios of at least one generator of the cylinder, but I don't know how to find that out from just the equation of the cylinder alone.

答:
The equation can be written $5(x+1)^2+5(y-1)^2+8z^2-2(x+1)(y-1)+8(y-1)z+8(x+1)z-6=0,$ and we can translate.

From @WillJagy's comment look at the plane $x+y-z=0.$ We're interested in the intersection $\langle 5x^2+5y^2+8z^2-2xy+8yz+8xz-6,x+y-z\rangle.$

Now rotate so that the normal vector $[\frac1{\sqrt{3}},\frac1{\sqrt{3}},-\frac1{\sqrt{3}}]$ goes to $[0,0,1],$ by way of the rotation matrix $\begin{pmatrix}\frac{1
 }{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\ -
 \frac{1}{\sqrt{2}\sqrt{3}}&-\frac{1}{\sqrt{2}
 \sqrt{3}}&-\frac{\sqrt{2}}{\sqrt{3}}\\ \frac{1
 }{\sqrt{3}}&\frac{1}{\sqrt{3}}&-
 \frac{1}{\sqrt{3}}\end{pmatrix}.$

This produces an equation in the new $XY$-plane: $6(X^2+2Y^2-1)=0,$ which has eccentricity $\frac1{\sqrt{2}}.$