问:
Hatcher Exercise 9, Chapter 1, using Borsuk-Ulam's theorem
Suppose $A_1,A_2,A_3$ are compact sets in $\mathbb{R}^3$, use Borsuk-Ulam theorem to show that there is one plane $P$ in $\mathbb{R}^3$ that simultaneously divides each $A_i$ into two pieces of equal measures.

I believe we should use the corollary of the theorem that says "whenever $S^2$ is expressed as the union of three closed sets $A_1,A_2,A_3$ then at least one of these sets must contain a pair of antipodal points," but I have no idea how to do it, any help is appreciated.

答:
BW's got the correct way, and here I add some details and mathematical rigor. $m$ is the Lebesgue measure on $\mathbb R^3$.

We can assume that $m(A_i)$ are all positive for $i=1,\,2,\,3$. If, say, $A_1$ and $A_2$ have measures $>0$ and $m(A_3)=0$, then we solve the problem for the three sets $A_1$, $A_1$ and $A_2$. We get a plane equally bisecting the measures of $A_1$ and $A_2$. And indeed this plane bisects the measure of $A_3$ equally: zero and zero.

So here's the outline (same as BW's but notations):

Let $v$ be an element of $S^2$, the unit sphere in $\mathbb R^3$. Let $t$ be a real number. Let $H^v_t$ be the set of all $x \in \mathbb R^3$ such that $x\cdot v < t$, where the dot means the Euclidean inner product.

Define $g_{i,\,v}(t)$ to be $m(H^v_t \cap A_i)/m(A)$. Then it is clear that $g_{i,\,v}$ is a monotonically increasing function $\mathbb R \to \mathbb R$. The compactness of $A_i$ ensures $g_{i,\,v}$ is real-valued. We will prove soon that $g_{i,\,v}$ is continuous, and that the set $g_{i,\,v}^{-1}(1/2)$ is a closed, bounded interval.

We define $f_i(v)$ to be the minimum of this interval $g_{i,\,v}^{-1}(1/2)$. Let $\hat f_i(v)$ denote the maximum of $g_{i,\,v}^{-1}(1/2)$. Generally, $f_i(v)$ and $\hat f_i(v)$ does not equal; think of the case that $A_1$ is a union of two disjoint closed balls. Anyway, $f_i$ and $\hat f_i$ are in fact continuous functions of $S^2$ into $\mathbb R$, which will be proved soon. Furthermore, we have $f_i(-v) = -\hat f_i (v)$ for all $v \in S^2$.

Finally, consider a function $h:S^2 \to \mathbb R^2$ given by $h = (f_2 - f_1 ,\, f_3 - f_1 )$. Use Borsuk-Ulam theorem with $h$ to find $v\in S^2$ for which $h(v)=h(-v)$, that is,

$$ f_2(v) - f_1(v) = f_2(-v) - f_1(-v) \,\,\,\, \text{and} \,\,\,\, f_3(v) - f_1(v) = f_3(-v) - f_1(-v). $$

By the relation between $f_i$ and $\hat f_i$, we know that

$$ f_1(v)+\hat f_1(v) = f_2(v)+\hat f_2(v) = f_3(v)+\hat f_3(v). $$

Let $c$ be the common value above. Then $c \in g_{1,\,v}^{-1}(1/2)\cap g_{2,\,v}^{-1}(1/2)\cap g_{3,\,v}^{-1}(1/2)$. Hence we conclude that the wanted plane is $x\cdot v = c$.

Now we dig into the details. For brevity, we drop the index $i$ from all expressions. Fix $v \in S^2$, and we show first that $g_v$ is continuous. The compact set $A$ is bounded in $\mathbb R^3$, so it is contained in a ball $B$ centered at $0\in \mathbb R^3$. Let $r$ be the radius of $B$. Let $C_v$ be the solid cylinder which is parallel to $v$ and tangent to $B$. Then $A$ is contained in $C_v$.

Let $\epsilon$ be a positive number. Choose a positive number $\delta$ less than $m(A)\epsilon / \pi r^2$. If two real numbers $t_1$ and $t_2$ satisfy $t_1 \le t_2$ and $t_2 -t_1 < \delta$, then we have

$$ g_v(t_2) -g_v(t_1) = m((H^v_{t_2} - H^v_{t_1}) \cap C_v \cap A) / m(A) \le \pi r^2 \delta / m(A) < \epsilon. $$

This shows that $g_v$ is uniformly continuous.

$A$ is compact, so if we choose a very large positive number $t$, then $g_v(-t)=0$ and $g_v(t)=1$. By the monotonicity and continuity of $g_v$, the set $g_v^{-1}(1/2)$ is closed, bounded, connected subset of $\mathbb R$: a closed interval.

Now we show that $f$ is continuous. Let $\theta$ be a real number between $0$ and $\pi/2$. Let $v=(v_1,\,v_2,\,v_3)$ and $w=(w_1,\,w_2,\,w_3)$ be elements of $S^2$ which satisfy $\cos ^{-1} (v\cdot w) = \theta$. We prove that

$$ f(v) \cos \theta - r \sin \theta \le f(w) \le f(v) \cos \theta + r \sin \theta, $$

which implies that if $w \to v$, then $f(w) \to f(v)$. Note that if $w$ tends to $v$, then $\theta$ tends to $0^+$.

We need to show the only case when $v$ is $(0,\,0,\,1)$. What is left is done by isometry of $\mathbb R^3$. Anyway in this case, $w_3 = \cos \theta$ and $\sin \theta = \sqrt{w_1^2 + w_2^2}$. Denote by $x=(x_1,\,x_2,\,x_3)$ an arbitrary point in $\mathbb R^3$. We have to show, in the cylindrical domain $x_1^2 + x_2^2 \le r^2$, that the plane $x_3 = f(v)$ lies between the planes $P^+_w$ and $P^-_w$, where

$$ P^\pm_w : \, w_1 x_1 + w_2 x_2 + w_3 x_3 = f(v) \cos \theta \pm r \sin \theta. $$

By the Cauchy-Schwarz inequality we have

$$ | w_1 x_1 + w_2 x_2 | \le \sqrt{w_1^2 + w_2^2}\sqrt{x_1^2 + x_2^2} \le r \sin \theta. $$

So the equation for $P^+_w$ implies $w_3 x_3 \ge f(v) \cos \theta$. $P^-_w$ implies the direction of inequality inverted. But by the range of $\theta$, the number $w_3 = \cos \theta$ is positive, therefore the desired result.

The continuity of $\hat f$ comes from the equation $f(-v) = -\hat f(v)$.


问:
Counting squares in a part of a graph. What is the correct answer?

Consider the following set of points in the $x - y$ plane:
$$
A = \{(a,\ b)\ |\ a,\  b\in \mathbb{Z}, \ 1\leqslant a\leqslant9\quad \text{and} \quad 1\leqslant b\leqslant 5\}.
$$
Find the number of squares whose vertices are points in $A$.

I reached an answer but don't know if it's right or not. (The source doesn't give the answer)

This reflects the essence of the approach.

First, we draw the blue line and choose any $2$ points on it and that will trace a square. The picture depicts $2$ cases. (The squares on the blue line)

Continuing this and advancing rightwards with every line before we reach the final yellow line, where one choice of points is absurd, because that square should go outside the figure itself. So, we have $6$ lines $5$ points long with one choice of points absurd so we need to exclude it.

After the $6\cdot\binom{5}{2} - 1$ squares we have, we only have to count squares in the last $3$ columns of points. Working on with the exact same reason and we have the lines in green, again with one choice of points absurd. So, there are $4\cdot\binom{3}{2}-1$ squares, totaling $$6\cdot\binom{5}{2} + 4\cdot\binom{3}{2} - 2\ \text{squares} = 70\ \text{squares}$$ but I doubt the number is too small.

What is the correct answer? How can it be reached? What will be the answer for the case when $0\leqslant a\leqslant9\ \text{and} \ 0\leqslant b\leqslant 5$? Thanks!

答:
I'm assuming the large rectangle as $R:=[0,n]\times[0,m]$ with $n\geq m\geq1$.

An axis-aligned lattice square $Q_r$ with side length $r\in\{1,2,\ldots,m\}$ has a certain mobility in $R$: It can be translated horizontally and vertically by integer amounts. It is easy to see that this $Q_r$ can be legally placed at $(n+1-r)(m+1-r)$ places as a lattice square in $R$.

Now such a $Q_r$ is in fact "housing" exactly $r$ lattice squares that have exactly the same mobility in $R$ as $Q_r$ has:

It follows that the total number of lattice squares that can be embedded in $R$ is given by
$$N=\sum_{r=1}^m r(n+1-r)(m+1-r)={m(m+1)(2nm+4n+2-m^2-m)\over12}\ ,$$
whereby the sum could be computed using the formulas for $\sum_{r=1}^m r^j$ $(0\leq j\leq3)$.

ContentLicense: CC BY-SA 4.0


问:
Order statistics and their CDF
Here's a problem that seems rather peculiar to me. This time I have no initial idea about how to solve it.

Let $X_1$, ..., $X_n$ be independent, real valued random variables with density $f$ and CDF $F$. Let $F_i$ denote the CDF of $X_{(i)}$.

a) What is the distribution of $F(X_{(i)})$ and $F_i(X_{(i)})$?

b) What is the variance of $F(X_{(i)})$?

[edit: I have realised by now that my approach was flawed and furthermore that my question needs clarification. So here is my second try.]

As I understand the question, we have real valued, i.i.d. random variables $X_1$, ..., $X_n$ with density $f$ and CDF $F$.

$X_{(1)}< ... < X_{(n)}$ are the corresponding order statistics with CDF $F_i:=F_{X_{(i)}}$.

I know the general formula for the CDF of order statistics. It is given by $$F_i(t)=\sum_{k=i}^n \binom{n}{k}F(t)^k(1-F(t))^{n-k}.$$

Now, the CDF of a random variable is a measurable function. Thus $F(X_{(i)})$ and $F_i(X_{(i)})$ are real valued random variables again. And a) asks for their distribution.

Per definition, we have $$F(t)=\mathbb{P}(X_i\leq t).$$ Hence $$F(X_{(i)})=\mathbb{P}(X_i\leq X_{(i)}).$$ This is the probability of the event that any $X_i$ is less or equal $X_{(i)}$. By definition of order statistics, we have $$F(X_{(n)})=\mathbb{P}(X_i\leq X_{(n)})=1,$$ as $X_{(n)}$ is the maximum of the $X_i$. But how do I derive the distribution of $F(X_{(i)})$ for $i \in \{1, ..., n-1\}$?

I think, if they were uniformly distributed, the answer would simply be $$F(X_{(i)})=i/n.$$ But their distribution ist unknown, so I'm stuck and have no idea how to proceed from here.

答:
Hints

If $X_{(i)}$ is the $i^{\text{th}}$ order statistic then it must be the case that $X_j \ge X_{(i)}$ for at most $i$ of the original random variables $X_j$.

There are ${n}\choose{i}$ possible ways of selecting which random variables obey the ordering in point 1 above.

Hopefully, the above two hints will get you started down the right track.

ContentLicense: CC BY-SA 3.0

Title: Order statistics and their CDF


问:
Can I use the mean value property for a function that is harmonic on only the interior of a ball?
Let $f$ be a continuous function on the closure of $U$ where $U=\{ (x,y) \in \mathbb{R}^2 : x^2+y^2<1 \}$ and harmonic on $U$,  If $$f(x,y)=x^2y^2 $$ on $\partial U$ I want to find $f(0,0)$.

Now I could of course solve the Laplace equation on the disk and then calculate the value at the origin, but I was wondering if I could apply the mean value property.

Now the mean value property holds if the function is harmonic on the closure of $U$, but here it is harmonic on only the interior, but I  actually tried to compute the mean value and I got the same value that I got when I solved the Laplace equation on the disk and calculated the value at the origin, so I am guessing that the mean value property must hold.

Perhaps the fact that $f$ is continuous on the closure is helpful.

My question is, can I use the mean value property in this case?

答: The mean value property holds for any disk of radius $1-\varepsilon$ with $\varepsilon$ as close to zero as you want. Observe that all those means over the boundaries are equal, independent of $\varepsilon$, since all of them are $f(0,0)$. By (uniform) continuity, the limit of the integrals over circumferences of radius $1-\varepsilon$ is equal to the integral over the circumference of radius $1$, so the property holds.


问:
Prove the two preimage sets have the same size under a surjective group homomorphism (for a finite group)
So I have a finite group $G$ and a surjective group homomorphism $\phi: G \to G'$. I was asked to show that for any $b,c\in G'$, $|\phi^{-1}(\{b\})|=|\phi^{-1}(\{c\})|$. As a hint, it told me to use the fact that $\phi(a)=\phi(b)$ iff $a^{-1}b\in\ker(\phi)$.

I first tried by choosing any $b_1,b_2\in\phi^{-1}(\{b\})\subseteq G$ and similarly for $c_1$ and $c_2$. Because the $b_i$'s are in the preimage of $b$, they map to the same element, i.e. $b$. Then I invoked the hint and I get $b_1^{-1}b_2\in\ker(\phi)$ and a similar result for the $c_i$'s. Now I am stuck and I have no clue how to proceed.

I thought of constructing a bijection between the preimage sets but I couldn't get anything out of it. Would using the left regular representation of $G$ help in anyway? I know it is supposed to feel something like a permutation of the group elements (and a permutation is a bijection). Would this idea prove the theorem if I apply it to the preimage sets?

Thanks in advance!

答:
By the homomorphism theorem, $G/\ker(\phi)$ is isomorphic to $G'$, as $\phi$ is surjective.

The isomorphism is given by $gH\mapsto \phi(g)$, where $H=\ker(\phi)$.

The preimage of $g'\in G'$ is $gH$ (left coset), where $\phi(g)=g'$.

All left cosets have the same cardinality by the bijection $H\rightarrow gH:h\mapsto gh$.


问:
Require help to solve an Inequality
$|0.022\cdot e^{-2t}\cdot sin(11t)|<0.01$

Values of $t$ for which the inequality holds.

Thanks in advance

答:
Taking into account @marty cohen's answer and using whole numbers, you want to find $t$ such that
$$f(t)=22\, e^{-2 t} \sin (11 t)-10<0$$ If you want a more precise result, you could use a numerical method for solving the equation.

The problem is that we need a reasonable guess. We have
$$f'(t)=22 e^{-2 t} (11 \cos (11 t)-2 \sin (11 t))$$ It cancels when
$$11 \cos (11 t)-2 \sin (11 t)=0 \implies t=\frac 1 {11}\tan ^{-1}\left(\frac{11}{2}\right)$$ but, by the second derivative test, this is a maximum. So, we need to start on the right of it. So , let us use
$$t_0=\frac 2 {11}\tan ^{-1}\left(\frac{11}{2}\right)$$ and Newton iterates are

$$\left(
\begin{array}{cc}
n & t_n \\
 0 & 0.252899 \\
 1 & 0.216374 \\
 2 & 0.215230 \\
 3 & 0.215227
\end{array}
\right)$$

The inequality then holds as soon as $t\geq 0.216$.

Title: Require help to solve an Inequality


问:
Uniformly distributed transformation
We let X be uniformly distributed  on the interval $(0,1)$. We must find the distribution of $Y=-log(X)$. I'm not sure how to do that but maybe I think maybe I can use that if X be uniformly distributed  on the interval $(a,b)$ then will the distribution be:
$$P(Y \leq y)=\frac{y-a}{b-a}$$, but how to use this and what if $P(Y > y)$?

答:
Slower: first, derive the CDF of $X=h^{-1}(Y)$. Since the function $h(X)$ is monotone decreasing, the sign changes:
$$
P(Y \leq y) = P(X \geq h^{-1}(y)) = 1 - e^{-y}=F_{Y}(y)
$$
This is because CDF of $X$ is $\frac{e^{-y}-0}{1-0}$. Hence PDF of $Y$ is
$$
f_{Y}(y) = e^{-y}
$$

Title: Uniformly distributed transformation


问:
What formula can be used to identify the number of points required in drawing desired line count?
I have to draw lines through a given points. Each point has to draw a line through other points. It draws only in one direction. So there is no duplicate lines passing through same points.

Now, I have been restricted to draw only 1000 lines. So I want to know how many points will I need to draw the stated count of lines? What is the formula that I can use to get the number of points required?

I thought of (n + 1) + (n + 2) + ( n + 3) + ..... = 1000. But don't know if this is right, as I have to put n = 0 to satisfy my equation.

Here is the illustration of what I am attempting:

Please let me know what I can do.

答:
It's the triangle numbers, in your diagram there are 10 lines, $\frac12n(n+1)$ with $n=4$, or use $\frac12n(n-1)$ with $n=5$.

Then to find out how many points are needed for 1000 lines, or the closest $n$ possible, form a quadratic equation and solve.


答: We have $$\phi^{-1}(b)=\{kb\mid k\in\ker(\phi)\}\\\phi^{-1}(c)=\{kc\mid k\in\ker(\phi)\}$$ so both sets have the same size. To see this let $x,y\in\phi^{-1}(b)$, then $$\phi(x)=b\\\phi(y)=b$$ so we have $\phi(xy^{-1})=e$, that is, $xy^{-1}\in\ker(\phi)$. Therefore $x=(xy^{-1})y$, so all elements differ by an element in the kernel.

Title: Prove the two preimage sets have the same size under a surjective group homomorphism (for a finite group)

Body:
So I have a finite group $G$ and a surjective group homomorphism $\phi: G \to G'$. I was asked to show that for any $b,c\in G'$, $|\phi^{-1}(\{b\})|=|\phi^{-1}(\{c\})|$. As a hint, it told me to use the fact that $\phi(a)=\phi(b)$ iff $a^{-1}b\in\ker(\phi)$.

I first tried by choosing any $b_1,b_2\in\phi^{-1}(\{b\})\subseteq G$ and similarly for $c_1$ and $c_2$. Because the $b_i$'s are in the preimage of $b$, they map to the same element, i.e. $b$. Then I invoked the hint and I get $b_1^{-1}b_2\in\ker(\phi)$ and a similar result for the $c_i$'s. Now I am stuck and I have no clue how to proceed.

I thought of constructing a bijection between the preimage sets but I couldn't get anything out of it. Would using the left regular representation of $G$ help in anyway? I know it is supposed to feel something like a permutation of the group elements (and a permutation is a bijection). Would this idea prove the theorem if I apply it to the preimage sets?

Thanks in advance!


问:
Orthogonally equivalent matrices
Suppose $A$ is any square matrix, then how do I show that there exists two orthogonal matrices $Q$ and $R$ such that $Q^TAR=A^T$?

I can show this when $A$ has distinct eigenvalues, but how do I show this for any square matrix $A$? I would really like someone to explain to me how $Q$ and $R$ act on the matrix.

In general, if $A$ and $B$ are two square matrices with the same eigenvalues, then when do have existence of two orthogonal matrices $Q_1$ and $Q_2$ such that $Q_1^TAQ_2=B$?

Let's assume the matrices to be real. Also if you have any reference materials on this, please share.

Thanks for the help.

答: Let $A=PU$ be a polar decomposition. Then $A^T=U^TP=U^TAU^T$. (Note that this is $U^TAU^T$, not $U^TAU$ or $UAU^T$. So, this is not a similarity transform.)

Title: Orthogonally equivalent matrices


问:
2nd Order Differential equation with Unit Step and initial conditions
I have some work but I can't get it to work quite yet, I need to solve this so I can plot it versus time.

The function is $90y''+30y'+1000(y-z(t))=0$,

$$z(t)= \begin{cases}0 & 0\leq t\leq5/9 \\ -\frac18 & 5/9<t\leq40/9\\0 & 40/9\leq t\leq 20/3\end{cases}$$
with $y(0)=0$ and $y'(0)=0$. I know $z(t)$ is a heaviside function but I am having difficulty moving forward. I've tried to simplify the main function to $90/1000y''+30/1000y'+y=z(t)$, and the heaviside notation I have is $z(t)=-\frac18u_{5/9}(t)$.

Help would be greatly appreciated. I apologize for the poor notation, I am new here and I do not know how to properly put in my functions.

答:
If you just want to plot it, you can also get the values numerically

```<code>def z(t): return -1/8 if 5<9*t<=40 else 0

t = np.linspace(0,7,500)
y = odeint(lambda y,t: [ y[1], -(3*y[1]+100*(y[0]-z(t)))/9 ], [0,0], t)
plt.plot(t,y)
</code>```

to get

Body:
If you just want to plot it, you can also get the values numerically

def z(t): return -1/8 if 5<9*t<=40 else 0

t = np.linspace(0,7,500)
y = odeint(lambda y,t: [ y[1], -(3*y[1]+100*(y[0]-z(t)))/9 ], [0,0], t)
plt.plot(t,y)

to get


问:
how to give elements of $A × B$ and $A ∩ B$
$A$ = {$3n+1: n∈Z$} and $B$ = {$12m+7: m∈Z$}

I have proven that $A⊄B$ by showing that $3n+1$ $⇒$ $3n+1+7-7$ ⇒ $3n-6+7$ ⇒ $3(n-2)+7$ ⇒ $4((3(n-2))+7)/4$ ⇒ $12(n-2)(1/4) +7$ ⇒ $12m-7$, where $m = (n-2)(1/4)$, and this proves that $A⊄B$, since $m$ will not always give an integer value.

I have also proved that $B⊆A$ by showing that $12m+7$ ⇒ $12m+7-6+6$ ⇒ $12m+6+1$ ⇒ $3(4m+2)+1$ ⇒ $3n+1$, where $n = (4m+2)$.

If my proofs for the above statements can be improved, let me know.

Now, I am trying to work on giving two concrete elements of $A × B$ and giving two concrete elements of $A ∩ B$.

$A × B$ is defined as $:=$ {$(a,b): a∈A$ and $b∈B$}

$A ∩ B$ is defined as $=$ {$x: (x∈A)$ and $(x∈B)$}

I am not exactly sure how to give concrete elements on $A × B$ and $A ∩ B$ or what it means to give concrete elements. If someone can explain this to me, I would really appreciate it.

Could I say that a concrete element of $A×B$ is $(1,6)$ and $(3, 10)$? I say that $b=6$ and $b=10$ because $(n−2)(1/4)$ will give an integer with those values of $b$? I am not sure what makes an element "concrete" and if I am understanding this correctly.

Could I say that a concrete element of $A∩B$ is $x=10$ or $x=6$ or $x=0$ or $x=14$. I am basically giving out numbers that $(n−2)(1/4)$ will yield an integer value for, since $B$ is contained in $A$, so all values of $B$ will work for $A$ but not the other way around?

Thanks for any answers given!

答:
$A\times B$ is the set of all ordered pairs or couples $(x,y)$ satisfying these 2 conditons :

(1) x belongs to $A$

(2) y belongs to $B$.

So , simply find an $x$ that meets the conditions to belong to $A$ and a $y$ that meets the conditions to belong to $B$, and "build" an ordered pair ( properly ordered of course!) with them; this ordered pair will automatcally be an element of $A\times B$.

In order to find  a member of $A\cap B$, just find an object (here , a number) that meets both the conditions to belong to $A$ and to belong to $B$ .

It could happen that no number meets these 2 conditions. In that case $ A\cap B$ would be identical to $\emptyset$ ( the empty set).

Now, in order to know whether it is the case, you could suppose you have found a number X that belongs to both sets.

You'd have $X = 3n+1$ and $X = 12m+7$.

This implies : $3n+1 = 12m+7$.

Some algebraic  manipulations will give you a certain relation between $n$ and $m$ that must occur in order $X$ to belong both to $A$ and to $B$.

Note : this relation must be compatible with the fact that $n$ and $m$ are integers.


问:
$A \subset R^n$ , that A open and volume and $f : A \to R$ a continuous no negative function that $f(x_0) > 0$ for some $x_0$. Prove $\int_{A} f > 0$
What I thought is using the theorem that prove an integrable and no negative function that $\int_{A} f = 0$ then the set {$x \in A$ : $f(x) \neq 0$} has measure zero. Apart from that I know that I can apply this theorem because a continuous function is integrable.

答:
If $f(x_0)>0$ and, and $x_0=(x_0^1,\ldots,x_0^n)\in A$, with $A$ open, since $f$ is continuous, there exists a $\delta>0$, such that
$$
B_\delta(x_0)=\{x\in\mathbb R^n: |x-x_0|<\delta\}\subset A,
$$
and
$$
|x-x_0|<\delta\quad\Longrightarrow\quad |f(x)-f(x_0)|<\frac{f(x_0)}{2}
$$
and hence
$$
|x-x_0|<\delta\quad\Longrightarrow\quad f(x)>\frac{f(x_0)}{2}.
$$
Now, since $f(x)\ge 0$ ,for all $x\in A$, then
$$
\int_A f\ge \int_{B_\delta(x_0)} f \ge \int_{B_\delta(x_0)} \frac{f(x_0)}{2} =
\frac{f(x_0)}{2}\cdot \mathrm{vol}\big(B_\delta(x_0)\big)>0.
$$


问:
Do I apply expectation correctly?
I stumbled upon the following problem recently:

A terrible crime has been committed.

The main attraction, a rare and precious diamond,
was stolen just before the museum opened its doors.
Patrons were horrified and the detectives were called immediately.

Here are their main suspects: (16 photos)

Alice needs to find the criminal by asking as few "yes"/"no" questions as possible.
How to achieve this might not be clear, but what about the opposite?

Suppose that Alice's strategy is to go through the suspects one by one, asking

"Was it Audrey?"
"Was it Bill?"
etc.
How many such questions would she have to ask, on average, to identify the criminal?

Is it correct to say?

$$
E(X) = S * P(X=x),$$ Let X represent the number of questions we have to ask,
$$E(X = i) = 1/16\\
E(X) = 1/16 * ( 1 + 2 + ... + 16)\\
E(X) = 1/16 * (16 * 17/2) = 8.5
$$

Solution from the author which I don't understand:

With this strategy for her questions,
Alice is committed to going through the suspects one by one,
asking Bob whether or not they're the criminal.

They could get lucky and identify the criminal on the first question,
or they could get unlucky and not do it until the last.
Since each suspect is equally likely,
we'd expect them to ask about 88 questions on average.

If they get through the first 15 suspects without finding the criminal,
then they know it's the person they haven't asked about.
For that reason, the average number of questions they'll need is 7.5,
slightly less than half the number of suspects.

答:
@trueblueanil is correct.
I will have the author's answer if I remove the 16th criminal prior to my calculations.</p>

Title: Do I apply expectation correctly?


问:
$\text{Hom}_k(k[x,x^{-1}],R)$ and $R^{\times}$ are bijective.

Let $R$ be a $k$-algebra where $k$ is a field. Then $\text{Hom}_k(k[x,x^{-1}],R)$ and $R^{\times}$ are bijective.

Originally, I want to describe a relationship between the set of $k$-algebra homomorphism $\varphi: k[x,x^{-1}]\to R$ and the set of units $R^{\times}$. So I guess they are bijective. I defined a map $\phi:\text{Hom}_k(k[x,x^{-1}],R)\to R^\times$ by $f\mapsto f(x)$. Then this map is well-defined as $f(x)$ has inverse $f(1/x)\in R$. And this map is injective as the $k$-algebra homomoprhism $f:k[x,x^{-1}]\to R$ is completely determined by $f(x)$. The problem is I wonder this map is surjective i.e. for given $a\in R^\times$, is there a $k$-algebra homomorphism $f:k[x,x^{-1}]\to R$ such that $f(x)= a$? Or there is another relation between those sets?

Note. All rings are assumed to be commutative with unity and I assumed if $f$ is a ring homomorphism then $f(1) = 1$.

Body:
Hint:
$k[x,x^{-1}]\cong k[x,y]/(xy-1)$. However,

$$\text{Hom}_k(k[x,y]/(xy-1), R)\cong \{(a,b)\in R^2: ab=1\}\cong R^\times.$$


问:
Angular excitations and Schrodinger operators with radial potential
Can someone please explain the following in mathematical language? What are angular excitations and s-waves mathematically?

"First of all, angular excitations only push the energy up, never down, so it is enough to analyze spherically symmetric s-waves...", c.f. first answer in
https://physics.stackexchange.com/questions/370901/solution-for-inverse-square-potential-in-d-3-spatial-dimensions-in-quantum-mec

Edit: change in the title.

答: When you solve the central force Schrodinger equation, you can use separation of variables, that is, you assume that the solution is a product of a term that is purely radial ( = dependent on the distance $r$ from the central force origin) and purely angular , i.e. dependent purely on the angles $\phi, \theta$. This gives eigen-solutions that have the form $\Psi=A(r)B(\phi, \theta)$. Once this is done, it is natural to consider solutions where the angular term is a constant, that is $\Psi = A(r)$. These are the S waves. "Angular excitations" are how the angular terms affect the solution eigenvalues.


问:
Prove that $\lim\limits_{n\to\infty}nx_n=0$
For a sequence $\left\{x_n\right\}$, prove that $\lim\limits_{n\to\infty}nx_n=0$, given that $\left\{x_n\right\}$ is decreasing and  $\sum_{n=0}^\infty x_n$ converges.

My given hint is to prove: $$\sum_{n=k}^\infty x_n \ge \sum_{n=k}^m x_n \ge \left(m-k\right)x_m$$
Then, I am meant to deduce: $$\limsup\limits_{m\to\infty}mx_m \le \sum_{n=k}^\infty x_n$$

How do I prove the latter deduction? I tried to write $m$ as $\left(m-k\right)+k$ to make $kx_m$ disappear but then I failed to make the $\limsup$ appear.

答:
I imagine, that $\sum_0^\infty x_n<\infty$ is also part of the assumption.

In that case,
$$
\sum_{n=0}^\infty\ge \sum_{k=\lfloor n/2\rfloor}^nx_k\ge (n-\lfloor n/2\rfloor+1)x_n\ge \frac{nx_n}{2.}
$$
It hence suffices to show that
$$
y_n=\sum_{k=\lfloor n/2\rfloor}^nx_k\to 0.
$$
If not, then as $y_n\ge 0$, there would be a $\varepsilon>0$ and  subsequence $y_{k_n}$, such that $y_{k_n}>\varepsilon$. Take a sub-subsequence $y_{j_n}$, such that,
$$
j_{n+1}>2j_n
$$
Then
$$
\sum_{n=1}^\infty\ge\sum_{k=1}^{j_n}x_k\ge \sum_{\ell=1}^n y_{j_\ell}\ge n\varepsilon\to\infty.
$$
Contradiction.


问:
Studying theorems in mathematics
Most math classes are 1. Learn the big concepts/definitions 2. Present theorems/corollaries 3. Prove them.
I presume it is essential to memorize the definitions and big ideas.

However, real mathematicians, do you use the theorems so much that you just memorized them all and they come to you as obvious, or do you understand the concepts so well that the theorems are naturally deduced when thinking about them and you don't think of them as "theorem 1, theorem 2, etc.", or do you actually have to refer to text when doing math? For myself, I know the main definitions - there is nothing to understand there, but when solving problems, I often search through the textbook looking for theorems that will be used to solve my problem. I feel that this is not a good method and the ultimate goal is to understand the concepts so well that the theorems are "obvious". I was wondering if this is indeed the case for mathematicians.

Lastly, how important is it to know the proofs of all the theorems? Is it important to be able to prove them without reference, or is it okay to simply take a theorem for granted once you have seen the proof once?

答:
I think it is really close to know the theorems and their proofs in the following sense.

When doing mathematics (whether it is an exercise sheet or research) you often want to prove things by reducing it to things you know. So it is helpful if not necessary to know the theorems of the subject you are working in and have an idea how and why they work. By this I don’t mean remembering proofs in full detail, but rather a sketch like extend a basis of the subspace to the whole space. Some theorems are crucial for a theory to work out (eg. existence of bases, extending bases and being able to define linear functions on a basis) and not knowing them is equivalent to not knowing the subject at all. But even if you cannot apply a (maybe less crucial) theorem per se, its proof may use methods/tricks applicable to your problem.

In fact it is useful to have this sort of knowledge of every bit of mathematics you learn. Great mathematics often comes by connecting seemingly different areas of mathematics. I mean what on earth has something rigid algebraic like group theory to do with analytic things like measure theory and Fourier-transformations? Some mathematicians knew enough about both topics to notice similarities of the objects/concepts/proofs and made a connection. These sort of connections allow new ideas to flow from one subject to the other, which is a really great thing.

There is a last kind of mathematical knowledge that is very common in research. It can become impossible to know each and every paper in and around your research topic, as new ones appear on a daily basis. But you can have an overview over the progress in the area in that by reading abstracts and attending conferences you are up to date in what progress and what connections were made. If you happen to need some statement you might recall that you read/heard something along the lines, locate and study it in more detail.

TLDR: As a student know the theorems and have a sketch in mind of how to prove them. You will need the ideas and methods later, sometimes even from different subjects combined. In research it is often more important to know that something is true than how it was proved.

Title: Studying theorems in mathematics


答:
Suppose you have $n$ points. Pick any point on the circle, and observe that $n-1$ lines emerges from that point. Also, every line is shared by $2$ points, so the total number of lines is $$\frac{n(n-1)}{2}$$

If you have $1000$ lines, then the minimum number of points is the smallest $n>0$ such that $1000\leq\frac{n(n-1)}{2}$. This can be solved to yield $$\left\lceil \frac{1+3\sqrt{889}}{2}\right\rceil=46$$

Title: What formula can be used to identify the number of points required in drawing desired line count?

Body:
I have to draw lines through a given points. Each point has to draw a line through other points. It draws only in one direction. So there is no duplicate lines passing through same points.

Now, I have been restricted to draw only 1000 lines. So I want to know how many points will I need to draw the stated count of lines? What is the formula that I can use to get the number of points required?

I thought of (n + 1) + (n + 2) + ( n + 3) + ..... = 1000. But don't know if this is right, as I have to put n = 0 to satisfy my equation.

Here is the illustration of what I am attempting:

Please let me know what I can do.


问:
Showing that $\mathbb{Q}(\sqrt{3+\sqrt{3}})$ is not Galois.
Why $\mathbb{Q}(\sqrt{3+\sqrt{3}})$ is not Galois?

I have only this:  Let   $\alpha=\sqrt{3+\sqrt{3}}.$

If $\alpha=\sqrt{3+\sqrt{3}}$ then $(\alpha^2-3)^2=3$ then $f(x)=x^4-6x^2+6$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ (Eisentein with the prime 2)

Therefore $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$

Now, to check that $[\mathbb{Q}(\alpha):\mathbb{Q}]\neq |aut(\mathbb{Q}(\alpha):\mathbb{Q})|$, is there a standard procedure to prove that the extension is not galois?

Actualization 1.
I got the following: If $\sqrt{3-\sqrt{3}}\in \mathbb{Q}(\alpha)$ then $\sqrt{6}\in \mathbb{Q}(\alpha)$. As also $\sqrt{3}\in\mathbb{Q}(\alpha)$ we have that $\sqrt{2}\in \mathbb{Q}(\alpha)$. Then $\mathbb{Q}(\sqrt{2},\sqrt{3})\subset \mathbb{Q}(\alpha).$
Because $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has the same degree over $\mathbb{Q}$ then $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\alpha)$.

Now, $\alpha=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$. I have this:

\begin{eqnarray} 3&=&a^2+2b^2+3c^2+6d^2\\
0&=&ab+3cd\\
1&=&6ac+4bd\\
0&=&ad+bc
\end{eqnarray}
and here I am still stuck

答:
Hint:

The roots of the minimal polynomial of $\;\sqrt{3+\sqrt3}\;$ are

$$\pm\sqrt{\frac{6\pm2\sqrt3}2}=\pm\sqrt{3\pm\sqrt3}$$

The question is thus: is $\;\sqrt{3-\sqrt3}\in\Bbb Q(\sqrt{3+\sqrt3})\;$ ? Because if it is then the signs in front of the square root don't make much difference: all the roots of the minimal polynomial would then be contained in the extension and thus this would be Galois...Try this way.


问:
Is every quadratic form in characteristic $2$ induced by a bilinear form?
Let $F$ be a field of characteristic $2$, and $q$ a quadratic form on $F^n$ for some positive integer $n$, i.e., $q:F^n\rightarrow F$ is a function such that $g(x,y):=q(x+y)-q(x)-q(y)$ is a bilinear form on $F^n$ and $q(cx)=c^2q(x)$ for all $c\in F$ and $x\in F^n$. Then does there exist a bilinear form $f$ on $F^n$ such that $q(x)=f(x,x)$ for all $x\in F^n$? If the answer is yes, how to find such an $f$?

答:
Yes, $f$ does exist. Keep your notation: If $(e_1,\ldots,e_n)$ is a basis of $F^n$, then let $f$ be the unique bilinear form such that $f(e_i,e_i)=q(e_i), f(e_i,e_j)=0$ if $i>j$ and $f(e_i,e_j)=g(e_i,e_j)
  $ if  $i<j$.

Then $$q(\sum_i x_i e_i)=\sum_iq(e_i)x_i^2+\sum_{i<j}g(e_i,e_j)x_ix_j=\sum_iq(e_i)x_i^2+\sum_{i<j}g(e_i,e_j)x_ix_j+\sum_{i>j}0\cdot x_ix_j=\sum_i f (e_i,e_i)x_i^2+\sum_{i<j}f(e_i,e_j)x_ix_j+\sum_{i>j}f(e_i,e_j)\cdot x_ix_j=f(\sum_i  x_ie_i, \sum_i x_i e_i).$$


问:
If $a,b\in \mathbb R^+$, $ |a-2b|\leq\frac {1}{\sqrt{a}}$, $|b-2a|\leq\frac {1}{\sqrt{b}}$ Prove $a+b\leq 2$

Question:  If $a,b\in \mathbb R^+$, $|a-2b|\leq\frac {1}{\sqrt{a}}$,  $|b-2a|\leq\frac {1}{\sqrt{b}},$ prove $a+b≤2$.

I figured out that $a+b\leq \frac {1}{\sqrt{a}}+\frac {1}{\sqrt{b}}$, but I am not sure how to prove that $a+b\leq 2$ after doing this.

Can anyone help me?

答:
Just my 2 cents to the Albus' answer, without using power mean. We can assume $0<a\leq b$ ($a=0$ can be checked easily) then $$\frac{(a+b)^2}{4}\leq a^3+b^3 \iff \frac{\left(1+\frac{b}{a}\right)^3}{4}\leq 1+\left(\frac{b}{a}\right)^3 \tag{1}$$
Now
$$f(x)=\frac{\left(1+x\right)^3}{4} -1-x^3=(1+x)\left(\frac{(1+x)^2}{4}-(1-x+x^2)\right)
=\\
(x+1)\left(\frac{-3+6x-3x^2}{4}\right)=-\frac{3}{4}(x+1)(x-1)^2$$
We can see that $f(x)\leq 0$ for $x\geq1$, which explains $(1)$.


问:
Find and classify critical points of $x^4-y^4-4xy^2-2x^2$, second derivative test is inconclusive.
As in the title I found critical points to be $(1,0)$-saddle point, $(-1,0)$- local minimum and $(0,0)$ is the one I have problem with.
Second derivative test is inconclusive in this case. If it was a saddle point I should find some curve in the graph that has inflection point at $(0,0)$ to prove it. Unfortunately, I suspect it to be local maximum and in this case I have no clue how can I prove that it grows in neighborhood of $(0,0)$ in every direction.

答:
The second-partial derivative test may leave us unedified when it is applied to functions which have terms involving power-functions with exponents larger than $ \ 2 \ . $  For the function here, the Hessian determinant function is

$$ \mathbf{H_f} \ \ = \ \ 4·(3x^2-1) \ · \ -4·(y^2 + 2x) \ - \ [ -8y ]^2 \ \ = \ \ -16·(3x^2-1) ·(y^2 + 2x) \ - \ 64y ^2 \ \  , $$

which equals zero at the origin and is inconvenient to "sort out" in the neighborhood.

Our experience with the second derivative test in single-variable calculus is that it is needful in such situations to investigate the properties of the function, and possibly the first-derivative function, in order to understand better what the behavior is in the vicinity of a point where the first two derivatives are zero.

This first graph is not u_sre's, but a graph of first partial-derivatives.  The curve in dark blue is $ \ f_x \ = 0 \ \rightarrow \ y^2 \ = \ x^3 - x \ , $ an elliptic curve. The curve in green is $ \ f_y \ = \ 0 \ \rightarrow \ y·(y^2+4x) \ , $ which is the union of the $ \ x-$ axis and the "leftward-opening" horizontal parabola $ \ y^2 \ =  -4x \ $ .  As expected, the intersections are the three points $ \ (-1,0) \ , \ (0,0) \ , \ $ and $ \ (1,0) \ . $  The meeting of the parabola and the elliptic curve at the origin, however, suggests that complications may be present there.

This second graph is u_sre's, shaded in green where our function is positive and in red where it is negative.  (The global minimum and the saddle point mentioned by Kran are marked as well.)  If a local extremum occurs at a point, it would be in a single region of either color (as we see with the saddle point $ \ (1,0) \ , $ the converse does not follow).  The origin, however, sits at the juncture of alternating "positive" and "negative" regions, indicating that the behavior of the function surface is not simple there.  This is a sign that the origin is a saddle point of the function.


问:
Estimate the number of sides of the die
Suppose you have a fair die with an unknown number of sides, $T$. Each side is distinct, marked with a complex symbol. Given enough rolls, can you estimate $T$?

My intuition is that can do this via a generalization of the mark and recapture procedure. Roll the die some large number of times, keeping track of all the sides that you have rolled and the number of times you have rolled them. Since the probability of rolling a previously rolled side in $N$ rolls is completely determined by $N$ and $T$, then expected proportion of unique to total sides $p$ should also be determined by $N$ and $T$. What is the relationship between $p$, $N$, and $T$?

As pointed out by Don Thousand, you can save yourself a lot of overhead and just track one arbitrarily picked side. Since the probability of rolling the chosen side on any one roll is $T^{-1}$, you expect that you would see the side repeated $NT^{-1}$ times. You can use standard statistical techniques to determine how precise this estimate likely is. If you have reasonably strong priors about the likely size of the population, within an order of magnitude or so, you can probably figure out how big of an $N$ you need to satisfy your requirements.

However, this procedure is inefficient, since we will likely have re-rolls of sides other than the one we have picked. The frequency of those re-rolls is almost equally as informative. How can we combine all the information we have to arrive at an estimate for $T$?

答:
Here is one way you could do it. Let $D_n$ be the number of distinct symbols observed after seeing $n$ rolls. Let $S$ be the number of symbols, and I'm assuming that we have prior distribution for $\mathbb{P}(S = s) = \pi(s)$ before we observe any dice rolls.

This answer shows that if $d \leq s$,
$$\mathbb{P}(D_n = d \mid S = s) = \frac{d!\binom{s}{d} \def\stir#1#2{\left\{{#1\atop#2}\right\}}\stir nd}{s^n},$$

where $\stir nd$ is a Stirling number of the second kind. Using Bayes' Theorem we can compute
$$\mathbb{P}(S = s \mid D_n = d) = \frac{\mathbb{P}(D_n = d \mid S = s) \mathbb{P}(S = s)}{\sum_s \mathbb{P}(D_n = d \mid S = s) \mathbb{P}(S = s)}.$$

To give an example, suppose I know that $S$ is in the set $\{6, 7, 8, 9, 10\}$ and that my prior is $\pi(6) = \pi(10) = 0.1$, $\pi(7) = \pi(9) = 0.2$, and so $\pi(8) = 0.4$. Now suppose we observe the following 15 rolls:

K K Q H X G A X X Q C C C G X

We have $D_{15} = 7$ because there are 7 distinct observed symbols: $\{A, C, G, H, K, Q, X\}$. We now have $\mathbb{P}(S = 6 \mid D_{15} = 7) = 0$ because we know there are at least 7 symbols. We now calculate the likelihood $\times$ prior terms:

$\mathbb{P}(D_{15} = 7 \mid S = 6) \mathbb{P}(S = 6) = 0$,

$\mathbb{P}(D_{15} = 7 \mid S = 7) \mathbb{P}(S = 7) = 0.433\ \cdot \ 0.2 = 0.087$,

$\mathbb{P}(D_{15} = 7 \mid S = 8) \mathbb{P}(S = 8) = 0.468\ \cdot \ 0.4 = 0.187$,

$\mathbb{P}(D_{15} = 7 \mid S = 9) \mathbb{P}(S = 9) = 0.360\ \cdot \ 0.2 = 0.072$,

$\mathbb{P}(D_{15} = 7 \mid S = 10) \mathbb{P}(S = 10) = 0.247\ \cdot \ 0.1 = 0.025$.

Then the posterior probabilities are

$\mathbb{P}(S = 6 \mid D_{15} = 7) = 0$,

$\mathbb{P}(S = 7 \mid D_{15} = 7) = 0.087/0.371 = 0.23$,

$\mathbb{P}(S = 8 \mid D_{15} = 7) = 0.187/0.371 = 0.50$,

$\mathbb{P}(S = 9 \mid D_{15} = 7) = 0.072/0.371 = 0.19$,

$\mathbb{P}(S = 10 \mid D_{15} = 7) = 0.025/0.371 = 0.07$.

So we are now even more convinced that $S=8$ than before (and I did actually use $S=8$ to generate the sequence of rolls).

You could also look into using Approximate Bayesian Computation (ABC) to solve this problem. There is a really nice example application of ABC here.

A related problem that you might find interesting if you haven't seen it before is the coupon collector's problem.

Title: Estimate the number of sides of the die


问:
Lower Limit Topology?
Show $(0,1)$ is open but not closed in the Lower Limit Topology.

I know that $[a,b)$ is open and closed in the lower limit topology, but I am not sure how to prove this one.

Thanks for any help.

答:
I'm going to attempt an answer to this based upon the other comments, hints and answers as part of an exercise in undertsanding the lower limit topology - I'd appreciate any pointers and comments.

I think any base of $\Bbb R_l$ contains subsets only of $\Bbb R$ of the form: $\{x:a\leq x<b:a,b\in\Bbb R\}$

The complement of $(0,1)$ in $\Bbb R$ is $(-\infty,0]\cup[1,\infty)$

$0$ is in the complement but by definition of the topology, every open set containing $0$ must have as a subset one of the form $[0,x):x>0$.  Since no such set is a subset of the complement, there is no open subset of the complement containing $0$ and therefore the complement is not open.  Therefore $(0,1)$ is not the complement of an open set and is therefore not closed.

Now to show it is open:

$$\bigcup_{0<x<1}[x,1)=(0,1)$$

Since $(0,1)$ is a union of open sets, it's open.

Title: Lower Limit Topology?


问:
Inequality $a^2+2b^2+8c^2\geq2a(b+2c)$
Can someone prove this inequality for the real numbers a,b,c?

$$a^2+2b^2+8c^2\geq2a(b+2c)$$

I have tried simple manipulation of the terms to get quadratic expressions, but since one cannot factor the $4ac$ on the right side, I abandoned that approach. Then I tried turning it into an expression where I could apply AM-GM, but that did not work either.
Maybe something like Muirhead, though I do not know where to start with that?
Any and all help would be greatly appreciated!!

Title: Inequality $a^2+2b^2+8c^2\geq2a(b+2c)$

Body:
Hints:

$$\frac12a^2 - 2ab + 2b^2 = \frac12(a-2b)^2 \ge 0$$

$$\frac 12 a^2 - 4ac + 8c^2 = \frac12(a-4c)^2 \ge 0$$


问:
Show that $\sum_{k=0}^N\binom{N}{k}x^k(1-x)^{N-k}(1-\alpha k/N)^{-1}$ is decreasing in $N$
I've come across the following sum of binomial coefficients:
$$S = \sum_{k = 0}^N \binom{N}{k}x^k (1-x)^{N-k} \frac{1}{1 - \alpha \frac{k}{N}},$$
where $x \in [0, 1]$ and $\alpha \in [0, 1)$.

I need to show $S$ is decreasing in $N$ (as the last term becomes smaller when $N$ increases), regardless of values of $x$ and $\alpha$ within the range specified above (that's what I obtained from simulation). I understand that $S$ ranges between $1$ and $\frac{1}{1 - \alpha}$ but don't know how to proceed further.

Much appreciated.

答:
It's sufficient to consider $0<x,\alpha<1$.
\begin{align*}
S&=\sum_{k=0}^N\binom{N}{k}x^k(1-x)^{N-k}\int_0^1 t^{-\alpha k/N}\,dt
\\&=\int_0^1\sum_{k=0}^N\binom{N}{k}(xt^{-\alpha/N})^k(1-x)^{N-k}\,dt
\\&=\int_0^1(1-x+xt^{-\alpha/N})^N\,dt=\int_0^1 t^{-\alpha f(x,t^{-\alpha/N}-1)}\,dt,
\end{align*}
where $f(x,y)=\log(1+xy)/\log(1+y)$ for $y>0$.

So we're left to show that $f(x,y)$ increases with $y$.

Here I'm going the "straightforward calculus" way. Denoting $f_y:=\partial f/\partial y$, $$\frac{f_y(x,y)}{f(x,y)}=\frac{x}{(1+xy)\log(1+xy)}-\frac{1}{(1+y)\log(1+y)}=\frac{g(xy)-g(y)}{y},$$ where $g(z)=\dfrac{z}{(1+z)\log(1+z)}$ is decreasing (for $z>0$), because $$\frac{g'(z)}{g(z)}=\frac{1}{1+z}\left(\frac1z-\frac1{\log(1+z)}\right)<0$$ since $0<\log(1+z)<z$. Hence $g(xy)>g(y)$, thus $f_y>0$ and we're done.


问:
What is the names of $A\vec{x}=\vec{b}$ linear equation system components?
Having  $A\vec{x}=\vec{b}$ .

What is the names of $A\vec{x}=\vec{b}$ linear equation system components?

答:
According to Elementary Linear Algebra by Venit and Bishop, we have:
$$A\vec{x}=\vec{b}$$
$A$ is the coefficient matrix of the system

$\vec{x}$ is the (column) vector of unknowns

$\vec{b}$ is the (column) of constants.

Granted, its probably not the most well-known book, but it's the one I have ;)


问:
Using ptolemy’s theorem i know that one diagonal is equal to the sum of the missing sides. What is the value of the missing sides?
cyclic quadrilateral containing an equilateral triangle. I know that using ptolemy’s theorem the one diagonal is equal to the sum of the two missing sides. What is the value of the missing sides?

答: Let the sides be $x$ and $(57-x)$. Apply cosine rule on the triangle with missing sides and solve for $x$.


Title: Inequality $a^2+2b^2+8c^2\geq2a(b+2c)$

Body:
Can someone prove this inequality for the real numbers a,b,c?

$$a^2+2b^2+8c^2\geq2a(b+2c)$$

I have tried simple manipulation of the terms to get quadratic expressions, but since one cannot factor the $4ac$ on the right side, I abandoned that approach. Then I tried turning it into an expression where I could apply AM-GM, but that did not work either.
Maybe something like Muirhead, though I do not know where to start with that?
Any and all help would be greatly appreciated!!

Body: $a^2 - a(2b+4c) + 2b^2+8c^2 \ge 0$ since $\triangle = (2b+4c)^2 -4(2b^2+8c^2) = -4b^2+16bc-16c^2 = -4(b-2c)^2 \le 0$.


Title: Inequality $a^2+2b^2+8c^2\geq2a(b+2c)$

Body:
$$a^2+2b^2+8c^2=\frac{1}{2}a^2+2b^2+\frac{1}{2}a^2+8c^2$$
Now By AM-GM :
$$\cases{\frac{1}{2}a^2+ 2b^2 \geq 2ab \\ \frac{1}{2}a^2+8c^2\geq4\sqrt{a^2c^2}=4ac}$$

Adding them together:
$$a^2+2b^2+8c^2 \geq2ab+4ac$$

And that’s a good place to stop.


答:
Your inequation is equivalent to

$$(a-b-2c)^2+(b-2c)^2 \ge 0 \tag{1}$$

The advantage of expression (1) is that it allows to clearly obtain the limit cases where there is an equality sign in (1) instead of a $">"$ symbol, i.e., iff

$$a=2b=4c$$

Edit: One can wonder how I have found expression (1). If you happen to know the concept of matrix associated with a quadratic form, here is the explanation:

$$a^2+2b^2+8c^2-2a(b+2c)$$

$$=\begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix}    1&   -1&-2\\
   -1&2&0\\
   -2&0&8 \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$$

which can be transformed, using the so-called (incomplete) Cholesky factorization:

$$=\begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix}    1&0&0\\
     -1&1& \ \ 0\\
    -2&-2 & \ \ 0   \end{pmatrix}\begin{pmatrix}    1&-1&-2\\
     \ \ 0&\ \ 1&-2\\
    0&0 &0   \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$$

$$=\begin{pmatrix}(a-b-2c)&(b-2c)&0\end{pmatrix}\begin{pmatrix}(a-b-2c)\\(b-2c)\\0\end{pmatrix}$$

$$=(a-b-2c)^2+(b-2c)^2$$

Title: Inequality $a^2+2b^2+8c^2\geq2a(b+2c)$


答:
$a,b,c$ are real numbers.
$$a^2+2b^2+8c^2\geq2a(b+2c)~~~~(1)$$
if for all real values of $a$
$$a^2-a(b+2c)^2+3b^2+8c^2\ge 0~~~~~(2)$$
$$\implies B^2 \le 4AC$$, as $A>0$, then
$$(b+2c)^2 \le 4(3b^2+8c^2) \implies -11b^2+4bc-28c^2\le 0$$
$$\implies 11b^2-4bc+28c^2\ge 0~~~~~(3)$$
In this quadratic $$D=B^2-4AC=4c^2-11.28 c^2 \le 0~~~~(4)$$
So (3) holds for all real values of $b$ and  (4) holds for all real values of $c$. Hence, this proves that (1), holds for all real values of $a,b,c$.

Title: Inequality $a^2+2b^2+8c^2\geq2a(b+2c)$


问:
Simple trigonometry question
I am just wondering how can you get from

$\cos(\pi t)=1/2 $    or $\cos(\pi t)=-1$

for $0<t<4$

to

t = 1/3, 1

t = 5/3, 7/3, 11/3 ,3

I got

$(\pi t) = \pi/3 +2k\pi $  and
$(\pi t) = 5\pi/3 + 2k\pi $

$t = 1/3 + 2k $   and $t = 5/3 +2k$

but i couldn't quite get t = 3, and t =1.... i'm not sure where i got it wrong... so could someone please explain this clearly to me?

答: $t = 1, 3$ look like the solution to $\cos(\pi t) = -1$, since $\cos(k\pi) =  -1$ for odd $k \in \mathbb{Z}$. Since you've restricted $0 < t < 4$, these are the two solutions.</p>

Title: Simple trigonometry question


问:
Finding eigenvectors of a $2 \times 2$ matrix
For the following matrix,

\begin{bmatrix}1&h\\0&1\end{bmatrix}

I was able to find the eigenvalues to be $1$ with multiplicity $2$. How to find the eigenvectors?

答:
If $h\neq0$, there is only one eigenvector (up to scaling). That's $(1,0)^T$. We must have $y=0$. We can see this from the equation
$$
\begin{bmatrix}1&h\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=1\cdot\begin{bmatrix}x\\y\end{bmatrix}\\
\cases{x+hy=x\\y=y}
$$
This is a common and important phenomenon, and as a consequence there are many ways of expressing it. For instance:

The algebraic multiplicity of the eigenvalue (you got $1$ as a double root of the characteristic polynomial) doesn't equal the geometric multiplicity (the eigenspace has dimension one)

Your matrix is not diagonalizable

Your matrix is defective

The minimal polynomial of your matrix has a multiple root


问:
Hard inequality in contest for regional Olympiad 2011
Let a,b,c strictly positive numbers.
Prove that
$$5(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+12 \geq 3(\sqrt{\frac{a+8b}{c}}+\sqrt{\frac{b+8c}{a}}+\sqrt{\frac{c+8a}{b}})$$

I have tried to use tangent method but seems useless. And I have tried Cauchy with no results...

答:
By Cauchy-Schwarz, we have $$\left(\sum\sqrt{\frac{a + 8b}c}\right)^2 \leq \left(\sum \frac b c \right)\left(\sum \frac {a + 8b} b\right).$$

If we write $X$ for $\sum\frac b c$ and $Y$ for $\sum\sqrt{\frac{a + 8b}c}$, then the above inequality translates to $$Y^2 \leq X(24 + X).$$

Therefore $$9Y^2 \leq 9X^2 + 216X \leq 9X^2 + 216X + (4X - 12)^2 = (5X + 12)^2.$$


问:
What is the largest number of k-element subsets of a given n-element set S such that...
Given a set S of n elements. What is the largest number of k-element subsets of S such that every pair of these subsets has at most one common element?</p>

答:
To every $k-$subset, associate the $k \choose 2$ subsets of pairs of elements. From the conditions in the question, every 2 $k-$subsets must have distinct pairs, hence there are at most $\frac{n\choose2} {k \choose 2} = \frac{n(n-1)}{k(k-1)}$ such sets. Note that this need not be an integer.

To get much better bounds, you will need to start to know what $n$ and $k$ are. For example, with $k=2$, we can get $\frac {n(n-1)}{2(1) }$ subsets easily. With $k> \frac{n}{2}$, we can only get 1 set, even though the above gives us a bound of $\approx 4$ if $k$ is close to $\frac{n}{2}$, and a bound of 1 if $k$ is close to $n$.


问:
Bounded set in finite normed vector space
Let $E$ be a finite-dimensional normed vector space and $A$ a non-empty and bounded subset of $E$.
Let $I = \{R\geq 0: \exists \mathcal{B}_{F}(x, R), A \subset \mathcal{B}_{F}(x, R)\}$, and $r=\inf I$.

Let $R_n = r+\frac{1}{n}$. Show that $\forall n \in \mathbb{N}^*$, there exists $x_n \in E$ such that $A \subset \mathcal{B}_F(x_n, R_n)$.

答: Your $I$ has the following property: if $s\in I$ and $t>s$ then $t\in I$. Therefore $I$ is an interval, more precisely either $(r,\infty)$ or $[r,\infty)$. Regardless, $r+\epsilon$ is always in $I$ for any $\epsilon>0$. Which, by the definition of $I$, means that there is $x\in e$ such that $A\subseteq B(x,r+\epsilon)$.


问:
Find all real numbers $x$ such that $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number
I want to find all values of $x\in \mathbb R$ such that the value of $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number.

I solved it as follows:

$x+2\sqrt{x}-1\ge 0$

$(\sqrt{x}+1)^2-2\ge 0$

$(\sqrt{x}+1)^2\ge 2$

$\sqrt{x}+1\ge 2$ or $\sqrt{x}+1\le -\sqrt{2}$

The second can't hold, from the first $x\ge 3-2\sqrt{2}$.

Similarly $(\sqrt{x}-1)^2\ge 2$, hence $x\ge 3+2\sqrt{2}$.

I find my solution to be very ugly. Is my solution correct and is there a neater approach?

答:
Let's use Mathematica for an overview plot:

```<code>ReImPlot[Sqrt[x + 2 Sqrt[x] - 1] + Sqrt[x - 2 Sqrt[x] - 1],
         {x, -1, 7}, PlotPoints -> 10^3,
         GridLines -> {{0, 3 - 2 Sqrt[2], 3 + 2 Sqrt[2]},
                       {0, 2 Sqrt[Sqrt[2] - 1], 2, 2 Sqrt[1 + Sqrt[2]]}},
         PlotTheme -> "Scientific"]
</code>```

For $x<0$ the function is purely real.

For $0\le x\le 3-2\sqrt{2}$ the function is purely imaginary.

For $3-2\sqrt{2}<x<3+2\sqrt{2}$ the function has nonvanishing real and imaginary parts.

For $x\ge3+2\sqrt{2}$ the function is purely real.

Body:
Let's use Mathematica for an overview plot:

ReImPlot[Sqrt[x + 2 Sqrt[x] - 1] + Sqrt[x - 2 Sqrt[x] - 1],
         {x, -1, 7}, PlotPoints -> 10^3,
         GridLines -> {{0, 3 - 2 Sqrt[2], 3 + 2 Sqrt[2]},
                       {0, 2 Sqrt[Sqrt[2] - 1], 2, 2 Sqrt[1 + Sqrt[2]]}},
         PlotTheme -> "Scientific"]

For $x<0$ the function is purely real.

For $0\le x\le 3-2\sqrt{2}$ the function is purely imaginary.

For $3-2\sqrt{2}<x<3+2\sqrt{2}$ the function has nonvanishing real and imaginary parts.

For $x\ge3+2\sqrt{2}$ the function is purely real.


问:
Curl of $u=(u_1,u_2)$
If I have $u=(u_1,u_2)$ is a vector of two components, then what is $\text{curl}\,u:=\nabla\times u$? where $\nabla=(\partial_x,\partial_y)$.

I think it should give a scalar and not a vector.

答:
The curl of a vector field is only really defined on vector fields $\mathbf{F}: \mathbb{R}^3 \to \mathbb{R}^3$. For a two-dimensional field $u = (u_1(x,y),u_2(x,y))$, the $z$ component is taken to be zero. The resulting curl has non-zero components only in the $z$-direction:

$$\nabla \times (u_1, u_2, 0) = \left(0, 0, \frac{\partial u_2}{\partial x} - \frac{\partial u_1}{\partial y}\right)$$

so people sometimes call the following scalar quantity the "curl":

$$\frac{\partial u_2}{\partial x} - \frac{\partial u_1}{\partial y}$$

Title: Curl of $u=(u_1,u_2)$


答:
The curl of a 2D vector can be defined with the following "intuition": The 2D vector $u=(u_1, u_2)$ can be "extended" to $\mathbb{R}^3$ as $\hat{u}=(u_1,u_2,0)$. Now we can take the "usual" curl of this $\hat{u}$ vector: $\hat{v}:=\nabla \times \hat{u}=(0, 0, v_3)$ and we identify $v_3$ with the curl of $u$, i.e.
$$\text{curl}(u_1, u_2)=\varepsilon_{321} \partial_2 u_1 + \varepsilon_{312}\partial_1 u_2 = \frac{\partial u_2}{\partial x} - \frac{\partial u_1}{\partial y}$$

Title: Curl of $u=(u_1,u_2)$

Body:
If I have $u=(u_1,u_2)$ is a vector of two components, then what is $\text{curl}\,u:=\nabla\times u$? where $\nabla=(\partial_x,\partial_y)$.

I think it should give a scalar and not a vector.


问:
Integral approximations.
Assume we have a general function $ \varphi $ such that $$ \intop_{a}^{b}|\varphi\left(t\right)|dt<\infty $$

It is well known that for any $\varepsilon>0 $ one can find a continuous function $ f $, such that $$ \intop_{a}^{b}|\varphi\left(t\right)-f\left(t\right)|dt<\varepsilon\, $$

I wonder it is possible to prove a stronger claim. Given $ \varepsilon>0 $ is it possible to find continuously differentiable $ f $ such that $$ \intop_{a}^{b}|\varphi\left(t\right)-f\left(t\right)|dt<\varepsilon\, $$ ?

If it is indeed possible, can we go further? is it possible for any $ \varepsilon>0 $ to find for any $n\in \mathbb{N} $ $f\in C^n((a,b)) $ such that $$ \intop_{a}^{b}|\varphi\left(t\right)-f\left(t\right)|dt<\varepsilon\, $$ ?

Thanks in advance.

答:
Yes, it is of course possible. The key words here are mollifier and Dirac-sequence. Meaning that this approximation can be achieved by convolution of a smooth so called kernel and $\varphi$.

This can be read in "Evans, Gariepy - Measure theory and fine properties of functions" (chapter 4.2) and "Alt - linear functional analysis". It is one of the classic results in approximation theory. Namely it is:

If $\Omega \subseteq \mathbb{R}^n$ is nonempty and open, then $C_0^\infty(\Omega)$, i.e. the $C^\infty(\Omega)$ functions with compact support in $\Omega$ are dense in $L^1(\Omega)$, the space of integrable functions on $\Omega$.

Title: Integral approximations.


问:
Sum separable partial differential equation over general domain
First off, thank you for taking your time out and considering my question. I just have a (rather) general question on partial differential equations.

So consider the following. Let $(x,y) \mapsto u(x,y)$ be some sufficiently smooth real-valued function over a rather general domain $\mathcal{F} \subset R^2$. And using subscripts to denote partial differentiation, suppose that,
$$
u_x(x,y) = g_1(x) \\
u_y(x,y) = g_2(y)
$$
In particular, note that $g_1$ is only a function of $x$ and $g_2$ is only a function of $y$. Now, if the domain $\mathcal{F}$ were rectangular in the form $[a,b] \times [c,d]$, then using the usual fundamental theorem of calculus, we would obtain $u(x,y) = \int_a^x g_1(\eta)d\eta + \int_c^y g_2(\eta) d\eta$.

But what about the case when $\mathcal{F}$ is rather general? That is, it could be connected, simply-connected, etc.? I can still see how $u(x,y)$ would have a similar separably additive form as above, but what about the region of integration? That is, do we have to consider, for each point $(x,y) \in \mathcal{F}$, the $x$-section $\mathcal{F}_y$ and $y$-section $\mathcal{F}_x$ of the domain $\mathcal{F}$ and integrate? Would it have the form like,
$$
u(x,y)
= \int_{\mathcal{F}_y} g_1(\eta) d\eta + \int_{\mathcal{F}_x} g_2(\eta) d\eta
$$
(and plus some constant of integration if it needs be)? And if it is indeed this form, to recover back the partial derivatives, we can't use the typical fundamental theorem of calculus, but we need to consider the Lebesgue differentiation theorem, correct?

Any help would be highly appreciated! Thank you!

PS. I'm aware that such sum-separable differential form can, via some change of variables, be rewritten into the wave equation. However, the traditional wave equation in $(t,x)$ where $x \in R^1$ allows for a very general consideration in the spatial coordinate $x$ but usually takes the time coordinate $t$ to be only the positive real line. But in my case above, this is not the case.

答:
Suppose $y$ still runs from $c$ to $d$, and for a fixed $y$, $x\in [a(y),b(y)]$, and for $y=c$ we have $x=a$. Integrate the first equation first
$$
u(x,y) = \int_{a(y)}^x g_1(\eta)d\eta + H(y) = G_1(x) - G_1(a(y)) + H(y),
$$
where $G_1' = g_1$. Now we take partial derivative with respect to $y$:
$$
g_2(y) = u_y(x,y) = -\partial_y\big(G_1(a(y))\big) + H_y = -g_1(a(y)) a_y(y) + H_y(y).
$$
Now we can integrate $y$ from $c$ for $H$:
$$
H(y) = \int_c^y g_2(\eta)d\eta + \int_c^y g_1(a(\eta)) a_{\eta}(\eta) d\eta = G_2(y) - G_2(c) + G_1(a(y)) - G_1(a),
$$
Plugging back to the first equation we have:
$$
u(x,y) = G_1(x) - G_1(a)+ G_2(y) -G_2(c)= \int_a^x g_1 + \int^y_c g_2.
$$
For general simply-connected domain in $\mathbb{R}^2$, we can approximate it by a sequence of collection of horizontal rectangles, and perform the integration one by one, then prove the convergence upon certain smoothness assumptions.</p>


问:
$(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$ if $ab+bc+ca=3$
Let $a,b,c$ be real positive number, $ab+bc+ca=3$.
Prove that $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$$
My attempt: $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 3\sqrt[3]{(a+b)(a+bc)(b+c)(b+ca)(c+a)(c+ab)}$$
$$P \ge 3\sqrt[3]{\frac{8}{9}\cdot(a+b+c)(ab+bc+ca)(a+bc)(b+ca)(c+ab)}$$
$$P \ge3\sqrt[3]{\frac{8}{3}\cdot(a+b+c)(a+bc)(b+ca)(c+ab)} $$
I can't prove $(a+bc)(b+ca)(c+ab)\ge 8$
Could you help me ?

答:
Your last inequlaity is equivalent to
$$
a^2b^2c^2+abc(a^2+b^2+c^2)+a^2b^2+b^2c^2+c^2a^2+abc\ge 8.
$$
Since
$$
a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=(a+b+c)^2-6~\text{and}~
\\
a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=9-2abc(a+b+c),
$$
the previous inequlaity is equivalent to
$$
a^2b^2c^2+abc(a+b+c)^2-6abc+9-2abc(a+b+c)+abc\ge 8,
$$
or
$$
a^2b^2c^2+abc((a+b+c)^2-2(a+b+c))+1\ge 5abc.
$$
Now note that $(a+b+c)^2\ge 3(ab+bc+ca)=9$, so $a+b+c\ge 3$ and hence,
$$
(a+b+c)^2-2(a+b+c)=(a+b+c-1)^2-1\ge 3.
$$
Thus, it's sufficient to prove that
$$
a^2b^2c^2+3abc+1\ge 5abc\Leftrightarrow (abc-1)^2\ge 0,
$$
which is true.


问:
Prove that a certain sum of binomial coefficients is divisible by a power of 2
As many will recognize the following expression is a closed form for the Fibonacci numbers. Can it be proved that

$$\frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{j} \quad \text{where} \quad n \in\mathbb{N}$$
are integers without using the fact that these are the Fibonacci numbers?

答:
Let
$$S_n=\frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{j} \quad \text{where} \quad n \in\mathbb{N}$$

By binomial theorem $$2\sum_{k=0}^{n/2}{n\choose 2k+1} (\sqrt{5})^{2k}=(1+x)^n-(1+x)^n$$
$$S_n=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n]=F_n$$
Where $F_n$ is Fibonacci number which is always a positive integer. Moreover, one may also check that $S_0=F_0=0$
and $S_1=1=F_1$.


答:
There are two values at which $\cos(\pi t)$ is a solution. You've made a dent in finding two of the four solutions to $$\cos(\pi t) = \frac 12, \;\;\text{for}\;\;t \in (0, 4).$$

There are four values at which $t$ gives a solution; they occur when $$0 \lt t \lt 4 \iff 0 \lt \theta \lt  4\pi$$

You found $t = \dfrac 13,\;\dfrac 53$, but also note that $t = \dfrac 73, \dfrac {11}{3} \implies \theta = \frac {7\pi}3, \theta = \frac{11}{\pi} \lt 4\pi = \pi t$

But you also found that $\cos(\pi t) = -1$ is also a solution. This happens when $t$ is an odd integer, so that $\pi t$ is an odd multiple of $\pi$:

$$\cos(\pi t) = -1 \implies \;\;t = 1, \;\;\text{or}\;\; t = 3, \quad0\lt t \lt 4$$

Title: Simple trigonometry question

Body:
I am just wondering how can you get from

$\cos(\pi t)=1/2 $    or $\cos(\pi t)=-1$

for $0<t<4$

to

t = 1/3, 1

t = 5/3, 7/3, 11/3 ,3

I got

$(\pi t) = \pi/3 +2k\pi $  and
$(\pi t) = 5\pi/3 + 2k\pi $

$t = 1/3 + 2k $   and $t = 5/3 +2k$

but i couldn't quite get t = 3, and t =1.... i'm not sure where i got it wrong... so could someone please explain this clearly to me?


问:
Fraction manipulation and binomial coefficients
I am working through a proof and I do not understand why the following equation holds:

$$\frac{m(m-1)(m+1)}{6} = \binom{m + 1}{3}$$

relative to the following definition of the binomial: $$\binom{n}{k} = \frac{n!}{k! \cdot (n - k)!} $$

What is the basic rule of fractions and multiplication justifying the first equality above?

答:
$${{m+1}\choose 3 }= \frac{(m+1)!}{3! (m-2)!}=\frac{(m+1)(m)(m-1) (m-2)!}{3!(m-2)!} =\frac{m(m-1)(m+1)}{6}$$

Second to third equality:

By definition , $$(m+1)! = (m+1)(m)(m-1)(m-2)\dots 2\cdot 1 \\ = (m+1)(m)(m-1) \big[ (m-2)(m-3) \dots 2\cdot 1\big] = (m+1)(m)(m-1) (m-2)!$$

Third to fourth equality:

I cancelled out the $(m-2)!$ from the numerator and denominator. This leaves $$\frac{(m+1)(m)(m-1)}{3!} $$ Now, $3!=6$.


问:
How to factorize $P=(X^2-4X+1)^2+(3X-5)^2$ in $\mathbb R[X]$?
First of all, I searched for roots. I knew that $\exists x\in\mathbb R,\, P(x)=0 \iff \exists x\in\mathbb R,\, x^2-4x+1 = 0 \text{ and } 3x-5 = 0$.

However, It is really easy to say that there is no real roots. So there is only complex non real roots.

And we know that $\deg{(P)}=4$ and $\text{dom}(P)=1$.

Which means that $P=(X^2+aX+b)(X^2+cX+d) = X^4+(a+c)X^3 + (ac+b+d) X^2 + (ad+bc) X+ bd $

With $a^2-4b<0$ , $c^2-4d<0$ and $a,b,c,d\in\mathbb R$ .

$P = (X^2-4X+1)^2+(3X-5)^2 = X^4 - 8 X^3 + 27 X^2 - 38 X + 26$. This gives us a system :

$$\left\{\begin{align}
a+c &= -8\\
ac+b+d&=27\\
ad+bc&=-38\\
bd&=26
\end{align}\right .
$$

However, I tried and i don't know how to solve this system.

I tried secondly with an euclidean division.

I found that $P = (X^2+aX+b)[X^2+(-8-a)X+(27-b+8a+a^2)]+(-38+8b+2ab-27a-8a^2-a^3)X+(-1+b^2-8ab-a^2b)$

With $a,b\in\mathbb R,\, a^2-4b <0$ and $(-8-a)^2-4(27-b+8a+a^2)<0$.

We have now this other system :

$$\left\{\begin{align}
-38+8b+2ab-27a-8a^2-a^3 &= 0\\
-1+b^2-8ab-a^2b&=0
\end{align}\right .
$$

However, I don't know how to solve this.

How would you do this ? And if it is the same thing as me, how could you solve those systems ?

答:
Assume (hope) that $a,b,c,d\in\Bbb Z$. If $bd=26$ then either:

$b=1,d=26$ or $b=26,d=1$ which gives $ac=0$ but this  is absurd;

$b=2,d=13$ which gives $13a+2c=-38$, equating with $a+c=-8$ gives $a=-2,c=-6$.

Indeed $(X^2-4X+1)^2+(3X-5)^2=(X^2-2X+2)(X^2-6X+13)$. This is the unique irreducible factorisation so we are done.


问:
Prove the two preimage sets have the same size under a surjective group homomorphism (for a finite group)
So I have a finite group $G$ and a surjective group homomorphism $\phi: G \to G'$. I was asked to show that for any $b,c\in G'$, $|\phi^{-1}(\{b\})|=|\phi^{-1}(\{c\})|$. As a hint, it told me to use the fact that $\phi(a)=\phi(b)$ iff $a^{-1}b\in\ker(\phi)$.

I first tried by choosing any $b_1,b_2\in\phi^{-1}(\{b\})\subseteq G$ and similarly for $c_1$ and $c_2$. Because the $b_i$'s are in the preimage of $b$, they map to the same element, i.e. $b$. Then I invoked the hint and I get $b_1^{-1}b_2\in\ker(\phi)$ and a similar result for the $c_i$'s. Now I am stuck and I have no clue how to proceed.

I thought of constructing a bijection between the preimage sets but I couldn't get anything out of it. Would using the left regular representation of $G$ help in anyway? I know it is supposed to feel something like a permutation of the group elements (and a permutation is a bijection). Would this idea prove the theorem if I apply it to the preimage sets?

Thanks in advance!

答: Pre-imaging establishes a partition of the domain for any map $f\colon X\to Y$, via the equivalence relation $x\sim x' \stackrel{(def.)}{\iff} f(x)=f(x')$. If $f$ is a group homomorphism, then the equivalence classes are the cosets of $\operatorname{ker}f$ (as hinted), which are all mutually equinumerous.


问:
GCD and LCM number theory problem
I am having some trouble finishing a solution to this problem, so I thought I might ask you all for help.

The problem goes as follows:
Find all natural numbers $a,b>0$ for which holds:

$$lcm(x,y)-gcd(x,y)=\frac{xy}{2021}$$

The problem is annoted with the hint, that 2021=43*47.

I have tried using the fact, that the $gcd=d$ from which follows $x=a*d$ and $y=b*d$. The $lcm(x,y)$ therefore must be $lcm(x,y)=a*b*d$.
Given this and the fact that the product of two numbers $x,y$ and their $gcd$ and $lcm$ is the same, meaning
$$xy=lcm(x,y)*gcd(x,y)=abd*d$$ I arrive at the equation:

$$(abd)-d=(abd^2)/2021$$
which equals
$$2021(ab-1)=abd$$

Where do I go from here? Any help or correction of mistakes that I may have made is greatly appreciated!

答: You rewrite it as $(2021 - d)ab = 2021$. Now use the factorization of $2021$ given in the hint.

Title: GCD and LCM number theory problem


答: You solved for $t$ s.t. $\cos\pi t=1/2$, but you also must solve for $\cos\pi t=-1$:$$\pi t=\pi+2\pi k\\\implies t=1+2k$$For $t\in(0,4)$ we find $t=1,3$ are solutions.</p>

Title: Simple trigonometry question


答: $r+\frac1n > r = \inf I$ we have that $r+\frac1n$ is not a lower bound for $I$, or $\inf I$ would not be the largest lower bound of $I$. So some $R_n \in I$ obeys $R_n < r+\frac1n$, and by the definition of $I$ this means there is some $x_n$ such that $A \subseteq B(x_n, R_n)\subseteq B(x_n, r + \frac1n)$ for this $R_n$. Done.

Title: Bounded set in finite normed vector space

Body:
Let $E$ be a finite-dimensional normed vector space and $A$ a non-empty and bounded subset of $E$.
Let $I = \{R\geq 0: \exists \mathcal{B}_{F}(x, R), A \subset \mathcal{B}_{F}(x, R)\}$, and $r=\inf I$.

Let $R_n = r+\frac{1}{n}$. Show that $\forall n \in \mathbb{N}^*$, there exists $x_n \in E$ such that $A \subset \mathcal{B}_F(x_n, R_n)$.


问:
Evaluating $\lim_{m \to \infty} \binom{m}{k} \cdot 2^{-m}$ for $0 \leq k \leq m$.
I am attempting to evaluate the limit $\lim_{m \to \infty} \binom{m}{k} \cdot 2^{-m}$ for $0 \leq k \leq m$, and I have been able to confirm via Mathematica that this limit is indeed 0. How would one actually go about proving this?

答: For simplicity let $m$ be even. $\binom{m}{k}$ is largest when $k=\frac12m$ [see here]. Hence $$\binom{m}{k}2^{-m}\le\frac{m!}{(m/2)!(m/2)!}2^{-m}.$$ Now using this you can find the limit to be zero.


问:
How to prove that $n\log(n) \in \mathcal{O}((n)^k)$?
I need to prove for  $k>1$ that is:

$n\log(n) \in \mathcal{O}(n^k)$

A also need to mention what happens in case that $k = 1$.
So actually, to prove
$n\log(n) \in \mathcal{O}(n)$

which I did so:

For big-Oh Notation $n\log(n) \in \mathcal{O}(n)$ exists an constant $c$ and $n_0$ such that for $n\gt n_0$ :

$n\log(n) \le  c * \mathcal{n}$  //  $\div\mathcal{n}$ gives:

$log(n) \le c$  // this is NOT TRUE weil there is no such a constant $c$ that for all $n\gt n_0$

But I have no idea how to prove $(n^k)$. Please help :)

答:
It suffices to prove that $\lim_{n \to \infty} \frac{n\log_2n}{n^k}$ exists. We have:
$$\lim_{n \to \infty} \frac{n\log_2n}{n^k} =\lim_{n \to \infty} \frac{\log_2n}{n^{k-1}} =  \lim_{n \to \infty} \frac{\frac{1}{ln(2)n}}{(k-1)n^{k-2}} = \lim_{n \to \infty} \frac{1}{ln(2)(k-1)n^{k-1}} = 0 $$ So the limit exists. Note that I have used the L'Hopital rule in the third equality and $k>1$.


问:
Infinitely Often iff divergent sum of probabilities

Given a probability space $(\Omega,\mathscr{F},P)$ and a sequence $A_1,\cdots \in \mathscr{F}$, show that $P(\lim\sup A_n) = 1 $ if and only if $\sum_{n=1}^\infty P(A_n|A)$ diverges for all $A$ with $P(A)>0$. Hint: Show that $P(A_n \ i.o)<1 \iff \sum_{n=1}^\infty P(A_n|A)<\infty$ for some $A$ with $P(A)>0$.

I tried proving the hint, for $\Rightarrow$ we take $B_m=\bigcup_{k=m}^\infty A_k$ from which, since $P(A_n \ i.o)<1$ then $P(B_m) \downarrow P(A_n \ i.o) < 1$ and therefore $P(B_m) < 1$ for some $m$, taking $A=B_m^c = \bigcap_{k=m}^\infty A_k^c$ we find
$$\sum_{n=1}^\infty P(A_n|B_m^c)\le \frac{1}{P(B_m^c)}(P(A_1)+\cdots P(A_{m-1})) < \infty $$
I tried proving $\Leftarrow$ then: Since $\sum P(A_n|A)$ converges then $\sum P(A_n \cap A)$ also does and therefore $P(\lim\sup (A_n\cap A)) = 0$ from Borel Cantelli's first lemma, I still can't jump from this to $P(\lim \sup A_n)<1$ since the only restriction on $A$ is that it has positive probability.

Edit: I changed the question so the partial solution is actually correct

The solution to $\Leftarrow$ is: $P(\lim\sup(A_n \cap A)) = P(\lim\sup(A_n) \cap A) = 0$ implies that $P(\lim\sup A_n) = P(\lim\sup (A_n) \cap A^c) \le P(A^c) < 1$

答:
Let $f=\sum I_{A_n}$, an extended real valued measurable function.

Then $f(\omega)=\infty$ iff $\omega \in \lim\sup A_n $.

If $P(\lim\sup A_n)=1$ then $f=\infty$ a.s. so $\int_A fdP=\infty$ for every set $A$ of positive probability. Hence, $\sum P(A \cap A_n)=\infty$ for every set $A$ of positive probability.

Conversely, if $\sum P(A \cap A_n)=\infty$ for every set $A$ of positive probability then $\int_A fdP=\infty$ for every set $A$ of positive probability. Taking $A=(f\leq n)$ we see that $P(f\leq n)=0$ for every $n$ so $f=\infty$ a.s., so $P(\lim\sup A_n)=1$.


问:
Probability Calculation for a Chess game

Here, I did not understand how to use the information. I know for the first 12 games the probability of Peter winning=6/12 and P(John winning)=4/12 and P(Draw)=2/12. But my question is about the next 3 games. So what am I supposed to do with the first information? How am I going to use that information in my calculation to solve these problems?

I still tried and considered three outcomes: Win, Lose, Draw. So total there would be 27 sample points. So for a. I did P=1/27. Please can somebody help me?

答:
From previous information, assuming that results of the previous matches hold for future matches, P(Peter wins) $=\frac6{12} = \frac1 2\;$ and so on

(a): P(Peter wins next $3$ games) $=(\frac1 2)^3 = \frac 1 8$

(b) If two games ended in a draw whose known Pr $= \left(\frac 1 6\right)$,

one game was a won by someone or other with $Pr =\left(\frac5 6\right)$

and thus indicated  $Pr = \binom 3 2 \left(\frac 1 6\right)^2\left(\frac 5 6\right)^1$

You should be able to do the rest similarly


问:
Change the order of a double integration
How should I change the order of the double integral of : $\int _0^1\:\int _{-x}^x 1\,dy\,dx$

答: The integration range is defined by$$0\le x\le1,\,-x\le y\le x,$$or equivalently$$-1\le y\le1,\,|y|\le x\le1,$$so the double integral is $$\int_{-1}^1\int_{|y|}^1dxdy=\int_{-1}^1(1-|y|)dy=2\int_0^1(1-y)dy=1,$$in agreement with the original $\int_0^12xdx=1$.


问:
Convergence in distribution and $L^1$-convergence of $f(X_n)$
Let $X_n$ and $X$ be independent random variables. If $X_n \to X$ weakly, then $E|f(X_n)-f(X)|\to 0$ for continuous bounded function $f$ ?

Because of convergence in distribution, there exist $Y_n,Y$ s.t.,$Y_n\to Y$ a.s., $X_n=Y_n$ and $X=Y$ in distribution. So, by Lebesgue's theorem $E|f(Y_n)-f(Y)|\to 0$.

But I don't know whether it holds that $E|f(X_n)-f(X)|=E|f(Y_n)-f(Y)|$.

Some one prove this property?

答: Not true. Let $(X_n)$ be i.i.d with standard normal distribution. Then $X_n \to X_1$ weakly and $-X_n \to X_1$ weakly (by symmetry of the distribution). If the result is true then $E|sin (X_n)-\sin (X_1)| \to 0$ and $E|-sin (X_n)-\sin (X_1)| \to 0$. By triangle inequality this gives $2 E|sin (X_n)| \to 0$. But $ E|sin (X_n)|= E|sin (X_1)|>0$.


问:
prove that $d(x,y):=\left|\frac x{1+y}-\frac y{1+x}\right|$ forms a metric on $(0,\infty)$
Let $X=(0,\infty)$. Define, $\displaystyle d(x,y):=\left|\frac x{1+y}-\frac y{1+x}\right|$. Show that $d$ forms a metric on $X$.

I have done all the conditions except triangle inequality. I've stuck here! Can anyone give some hint for triangle inequality ?

We have,

\begin{align*}
d(x,z)&=\left|\frac x{1+y}-\frac y{1+x}\right|\\
&=\left|\frac{(x-z)(1+x+z)}{(1+x)(1+z)}\right|\\
&\le \left|\frac{(x-z)(1+x+z+xz+y+y^2)}{(1+x)(1+y)(1+z)}\right| \tag 1
\end{align*}
If we can write the inequality (1) then I can prove the result. But how to prove that $\displaystyle \frac{(1+x+z+xz+y+y^2)}{(1+y)}\ge 1+x+z$ ?

Please help me in proving this , otherwise any easiest way for triangle inequality ?

Body: It's not a metric. $d(1,3)>d(1,2)+d(2,3)$.


问:
show that sequence of functions $f_n(x) = (nx^2+1)/(nx+1)$ converges uniformly on $[1,2]$
I have found out that this sequence of functions approaches to $1$ at $x=1$, approaches to $2$ at $x=2$, approaches to $x$ when $1<x<2$.
Now to prove the sequence converges uniformly in $[1,2]$, what should be N ?? And what is the trick to find N ?

答: If $x\in[1,2]$, then\begin{align}|f_n(x)-x|&=\left|\frac{nx^2+1}{nx+1}-x\right|\\&=\left|\frac{1-x}{nx+1}\right|\\&\leqslant\frac1n.\end{align}Can you take it from here?


问:
Special permutations of $\{1,2,3,\ldots,n\}$
How do you show that number of permutations of $\{1,2,3,\ldots,n\}$ such that image of no two consecutive numbers is consecutive is

$$n! + \sum_{k = 1}^{n}(-1)^k\sum_{i = 1}^{k}\dbinom{k - 1}{i - 1}\dbinom{n - k}{i}2^i(n - k)!$$

In short we need to find number of permutations of $\{1,2,3,\ldots,n\}$ such that none of the following occur: $12, 23, \ldots, (n-1)n \quad $ and $ \quad21, 32, \ldots, n(n-1)$ that is no adjacent numbers should be consecutive.

I tried proving the formula but didn't get any satisfactory result, It seems to be inclusion exclusion principle would work, but there are too many cases to count. I tried to find a recurrence relation, but couldn't do it either. Afterwards I tried to get a generating function for the same but didn't succeed. I don't see any other approach to get through this but I think the most useful tool would be PIE however I'm not finding a good way to use PIE since number of cases are too much. Any help or hint would be highly appreciated. Thanks!

Note that:

I have read almost all the references related to the problem from OEIS.

I have read the whole paper https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-38/issue-4/Permutations-without-Rising-or-Falling-omega-Sequences/10.1214/aoms/1177698793.full   but in the paper there aren't any rigorous proofs and most of the proofs are just excluded simply by saying that 'use basic PIE to derive this'. I am looking for a more direct poof using enumerative combinatorics or generating functions.

I highly appreciate your time and efforts. Thanks.

答:
Here's a sign-reversing involution argument giving Flajolet's generating function. The argument and terminology are my own.

An "underlined permutation" is a permutation where each element is underlined, and consecutively increasing or consecutively decreasing adjacent subsequences may be underlined together. For instance, $\underline{3}\,\underline{1\,2}\,\underline{9\,10}\, \underline{8}\, \underline{7\,6\,5}\,\underline{4}$ is an underlined permutation.

There is a reasonably natural involution $\phi$ on the set of underlined permutations: take the rightmost adjacent increasing or decreasing pair and either break the underline between them or combine their underlines.

For example,
\begin{align*}
\phi(\underline{3}\,\underline{1\,2}\,\underline{9\,10}\, \underline{8}\, \underline{7\,6\,5}\,\underline{4})
&= \underline{3}\,\underline{1\,2}\,\underline{9\,10}\, \underline{8}\, \underline{7\,6\,5\,4} \\
\phi(\underline{3}\,\underline{1\,2}\,\underline{9\,10}\, \underline{8}\, \underline{7\,6\,5\,4})
&= \underline{3}\,\underline{1\,2}\,\underline{9\,10}\, \underline{8}\, \underline{7\,6\,5}\,\underline{4}
\end{align*}

Of course, if there are no adjacent increasing or decreasing pairs, we can do nothing, so define $\phi$ to be the identity in that "degenerate" case. In particular, you want to count the number of fixed points of $\phi$.

We may construct an underlined permutation in three steps: after writing $1\,2\,\ldots\,n$ in a line, (1) underline consecutive sublines together, (2) optionally reverse the order of individual sublines, and (3) permute the underlined groupings arbitrarily. For example, we might have

$\underline{1\,2}\,\underline{3}\,\underline{4}\,\underline{5\,6\,7}\,\underline{8}\,\underline{9\,10}$

$\underline{1\,2}\,\underline{3}\,\underline{4}\,\underline{7\,6\,5}\,\underline{8}\,\underline{9\,10}$

$\underline{3}\,\underline{1\,2}\,\underline{9\,10}\, \underline{8}\, \underline{7\,6\,5}\,\underline{4}$

Define two statistics on an underlined permutation: the number $N$ of elements being permuted and the number $n$ of underlined groupings. The compositional formula for bivariate ordinary generating functions gives

$$\sum x^{N(\pi)} q^{n(\pi)} = \sum_{n=0}^\infty n! q^n (x+2x^2+2x^3+2x^4+\cdots)^n = \sum_{n=0}^\infty n! (qx)^n \frac{(1+x)^n}{(1-x)^n}\label{*}\tag{*}$$

where the sum is over all underlined permutations $\pi$.

Plug in $q=-1$ to $\eqref{*}$. The left-hand side becomes the generating function for the difference of the number of underlined permutations on $N$ elements with an even $n$ minus the number with odd $n$. Since $\phi$ preserves $N$ and, in the non-degenerate case, increments or decrements the number $n$ of underlined groupings by $1$, only degenerate $\pi'$ counted by A002464 contribute. Note that $N(\pi') = n(\pi')$ since every element must be underlined separately. Thus

$$\sum_{\pi'\text{ in }A002464} (-1)^{N(\pi')} x^{N(\pi')} = \sum_{n=0}^\infty n! (-x)^n \frac{(1+x)^n}{(1-x)^n}.\label{**}\tag{**}$$

Apply $x \mapsto -x$ to $\eqref{**}$ to get Flajolet's generating function:
$$\sum_{\pi'\text{ in }A002464} x^{N(\pi')} = \sum_{n=0}^\infty n! x^n \frac{(1-x)^n}{(1+x)^n}.$$

Edit: After writing this, I noticed Flajolet and Sedgewick discuss this generating function briefly and somewhat vaguely on p.373 of their text (which was mentioned on OEIS). They give a sketch of a sketch, but the ideas are similar.

Some routine but tedious calculations do allow you to extract the double sum formula above from this generating function. I probably don't have enough interest to type those manipulations out.


问:
When does $E[max(X_1, ..., X_n)]$ = $\sum_{i=1}^n E[X_i]$ hold true?
Consider independent random variables $X_1, ..., X_n \ge 0$ with $E[X_i] > 0 $.

In which cases does $E[max(X_1, ..., X_n)]$ = $\sum_{i=1}^n E[X_i]$ hold true?

For two Variables I found $X_1 = 1$ and $X_2 = 0$ with probability $1-\epsilon$ and $\frac{1}{\epsilon}$ otherwise when $\epsilon$ approaches $0$.

Is there an example for more than two variables?

答: Hint: $\max (X_1,X_2,...,X_n)\leq X_1+X_2+..+X_n$. If the given condition holds then both sides have the same expectation. This is true iff $\max (X_1,X_2,...,X_n)= X_1+X_2+..+X_n$ with probability $1$  which mean all but one $X_i$ is $0$ with  probability $1$. Can you now use independence to show  that there exist $i$ such that $P(X_j=0 \forall j \neq i)=1$? [Look at $EX_iX_j=EX_iEX_j$].


答:
Hint: write it as $\,f(a)=0\,$ so $\,a\mid f(0)\,$ since $\,f\,$ is a polynomial in $\,a\,$ with integer coef's. Thus the integer values that $a$ can take are constrained to the finite number of factors of $\,f(0)\,$ (if $\,f(0)\neq 0)$

Remark $ $ This can be viewed as special (trivial) case of the ubiquitous Rational Root Test, i,e., an integer root of $f$ divides its constant term.

Title: GCD and LCM number theory problem

Body:
I am having some trouble finishing a solution to this problem, so I thought I might ask you all for help.

The problem goes as follows:
Find all natural numbers $a,b>0$ for which holds:

$$lcm(x,y)-gcd(x,y)=\frac{xy}{2021}$$

The problem is annoted with the hint, that 2021=43*47.

I have tried using the fact, that the $gcd=d$ from which follows $x=a*d$ and $y=b*d$. The $lcm(x,y)$ therefore must be $lcm(x,y)=a*b*d$.
Given this and the fact that the product of two numbers $x,y$ and their $gcd$ and $lcm$ is the same, meaning
$$xy=lcm(x,y)*gcd(x,y)=abd*d$$ I arrive at the equation:

$$(abd)-d=(abd^2)/2021$$
which equals
$$2021(ab-1)=abd$$

Where do I go from here? Any help or correction of mistakes that I may have made is greatly appreciated!


问:
The motivation behind the properties of height functions
I'm reading Hindry & Silverman's book on Diophantine Geometry, specifically the section on height functions.

At the beginning they speak of two important properties that a height function $h:V\rightarrow [0,\infty)$ should satisfy: first, the set of points in the complex projective variety $V$ of bounded height should be finite. Second, it should translate geometric properties of points in $V$ to arithmetic properties.

The second property is a natural thing to strive for, if somewhat difficult to quantify. But I don't see any particular reason to ask for the first property to be true. I understand that such a property is fundamental in many finiteness proofs - to show that $V(\mathbb{Q})$ is finite, for example, one can simply show that its points have finite, bounded height.

So, what exactly is the motivation for seeking this property in height functions? Or was the theory actually developed the other way around - geometers knew that they needed functions with this property in order to tackle certain problems?

答:
I think you are "putting the cart before the horse" a little here. Height functions were invented as a tool for finiteness / finite-generation proofs. Their pedigree goes all the way back to Fermat's proof that $x^4 + y^4 = z^4$ has no nontrivial integer solutions, which is fundamentally an induction argument: assuming the existence of a solution, you produce a second solution which is strictly "smaller", in some natural sense, and then you deduce a contradiction because there must be a "smallest" solution.

The Mordell--Weil theorem (finite generation of $E(\mathbf{Q})$ for $E$ an elliptic curve) uses a similar idea, and the naive height of points is used to quantify what it means to be "smaller". The general theory of height functions essentially developed out of systematizing and axiomatizing what properties a function would need to have in order to make that argument work, and here the finiteness of the set of points of height $\le X$ is crucial. So I think that it would be entirely unreasonable to use the word "height" for any function not having this key finiteness property.


问:
Group operator vs. complex number multiplication operator in the multiplicative property of a character of a group
A character $f$ of a group $G$ is defined as a complex-valued function defined on $G$ that has the multiplicative property $f(ab) = f(a)f(b)$ for all $a, b$ in $G$, and if $f(c) \ne 0$ for some $c$ in G.

I want to make sure I understand the definition of character right. I am going to make some claims per my understanding. Please tell me if my claims are true or false.

The group $G$ need not be a group of complex numbers. It may be a group of vectors, matrices or any other elements that satisfy the group postulates.

The domain of $f$ need not contain complex numbers.

The codomain of $f$ contains complex numbers.

In the notation $f(a)$, $a$ is an element that belongs to $G$ and the value of $f(a)$ is a complex number.

The notation $f(ab)$ denotes the function $f$ applied to $a \circ b$ where $ \circ $ represents the group operator, i.e. if $ \circ $ is $ + $, then $f(ab)$ denotes $f(a + b)$.

The notation $f(a)f(b)$ denotes complex number multiplication of $f(a)$ and $f(b)$.

Let $ a \in G $. Let $ \circ $ be the group operator. Let $ \times $ represent the multiplication operator to multiply complex numbers. The multiplicative property of the character can now be stated as $f(a \circ b) = f(a) \times f(b)$.

Are all these claims true or do you find some of them false?

答:
Your statements are correct with one proviso. As the Wikipedia article
Character theory states:

Let $V$ be a finite-dimensional vector space over a field $F$ and let
$\rho:G\to \text{GL}(V)$ be a representation of a group $G$ on $V$. The
character of $\rho$ is the function $\chi_\rho:G\to F$ given by
$\chi_\rho(g) = \text{Tr}(\rho(g))$ where Tr is the trace.

It later states

A character of degree $1$ is called linear.

And also

However, the character is not a group homomorphism in general.

To summarize, your $\,f:G\to \mathbb{C^\times}\,$ is a group homomorphism
which is of degree $1$ which is linear  and corresponds to $\rho$. Since
it is linear, then $\rho=\chi_\rho$ which is its own character. In general,
a character can be equal to $0$, but linear characters can not take the
value $0$.
Your definition of character seems to be confusing the representation
$\rho$ with its character $\chi_\rho$ which is easy to do if the degree is
$1$.

For closely related but slightly different kind of characters compare
group characters with the Wikipedia article on
Dirichlet character
defined on the multiplicative group of units of integers modulo $\,n.$


问:
How to find the minimal polynomial?
Let

$$T : \mathbb{R}^{n \times n} \to \mathbb{R}^{n \times n}, \qquad T(A) = A^{t}$$

I want to find its minimal polynomial, but in order to do that I need to find the matrix basis $[T]_{\tilde{e}}$ lets say $\tilde{e}$ is the basis of a matrix. Which means that its consider by the vectors after we bring the matrix to its  row-echelon form.

I am quite confused could someone help me understand how to find $[T]_{\tilde{e}}$ and the minimal polynomial?

答: You don't need to pick any basis. Since $(A^t)^t$ is identically equal to $A$, $p(x)=x^2-1$ annihilates $T$. If you can prove that no linear polynomial annihilates $T$ (note: whether $n$ is greater than $1$ is essential here), then $p$ is the minimal polynomial of $T$.

Title: How to find the minimal polynomial?


问:
Proof that the tangent space to an open submanifold can be identified with the tangent space to the whole manifold
I have one question about the proof given by John M. Lee in his book. Please find the sentence underlined with red in the following picture. How could Prof. Lee guarantee that such $B$ does exist? Thank you.

P.S. If I am not mistaken, $\overline{B}$ is the closure of $B$ in the space $M$.

答:
The short answer is that each (topological) manifolds is a regular topological space.

To see this, let $p \in M$ and let $U$ be an open neigbhorhood of $p$ in $M$. Choose an open $W \subset M$ such that $p \in W \subset U$ and $W$ is homeomorphic to an open subset $W' \subset \mathbb R^n$. Let $h : W \to W'$ be a homeomorphism. There exists an open ball $B_r(h(p))$ around $h(p)$ with radius $r > 0$ such that $B_r(h(p)) \subset W'$. Then $B' = B_{r/2}(h(p))$ has the property that $\overline{B'}$ is compact and $\overline{B'} \subset W'$. Now take $B = h^{-1}(B')$. This is an open subset of $W$, hence also an open subset of $M$. Note that $h^{-1}( \overline{B'})$ contains $B$ and is compact, hence closed in $M$ since $M$ is Hausdorff. Thus

$$p \in B' \subset \overline{B} \subset h^{-1}( \overline{B'}) \subset W \subset U .$$


问:
How to solve these type of differential equations: $y''+1/(1+x)y'-(1/(1+x)^2)y=0$
Is there any particular way one can go about solving such questions. Here, I assumed $P=1/(x+1)$, $Q=-1/(x+1)^2$ and tried solving it but to no avail. Thanks in advance.

答: This is an Euler-Cauchy DE around the singular point $x=-1$ and can be solved as such making the trial $y=(1+x)^m$.


问:
Taylor Series of $ f(x) = \frac{5}{3-6x} $
I am trying to get the Taylor series at $x_0 = 0 $ of $ f(x) = \frac{5}{3-6x} $. I thought about using the modified geometric series $  \sum_{k=0}^{\infty} 5(-6x)^{k} $ . But I guess that isn't quite right because $\mid q \mid=(-6x)$ might be $ \geq 1 $ and I am missing the 3 in the denominator.

Thanks.

答: Note that$$f(x)=\frac53\times\frac1{1-2x}=\frac53\sum_{n=0}^\infty2^nx^n$$if $|x|<\frac12$. But$$\frac53\sum_{n=0}^\infty2^nx^n=\sum_{n=0}^\infty\frac{5\times 2^n}3x^n$$and this is the Taylor expansion of $f$ centered at $0$.


问:
Normed Space $L^2$ and Vector Space $l^2$
I am really confused about $L^2$-Space and $l^2$-Space. Can anybody explain what is the difference between the $L^p$ and $l^p$ space?

Also in $L^p$-Spaces, we see that $L^2 \subset L^1$ (for $L^2[(0,T_f)], T_f >0 $). On the other hand for $l^p$-Spaces, we see that $l^1 \subseteq l^2$.

How can we show that $l^1 \subseteq l^2$?

I don't understand why(/how) does the subset sign change for the two different cases. Thanks in advance.

答:
Let $\mu$ be a measure on some measurable space $X$. You may then define:

$$L^p(X,\mu):= \{f:X\to\Bbb C\,\mid f\text{ measurable and }\int_X|f|^pd\mu<\infty\}$$

For example if $X=\Bbb N$ and $\mu_c$ is the counting measure then a function in $L^p(\Bbb N, \mu_c)$ is just a sequence $f(n)$ satisfying $\sum_{n\in\Bbb N} |f(n)|^p<\infty$, ie this is $\ell^p(\Bbb N)$. Similarly if $X$ is some interval in $\Bbb R$ and $\mu$ is the Lebesgue measure you recover the usual $L^p$ space.

With this perspective whether or not you get some kind of an inclusion $L^p(X,\mu)\subseteq L^q(X,\mu)$ is a property of the measure space $(X,\mu)$.

Definition (not standard names)

Call a measure space $(X,\mu)$ IR-divergent if $\mu(X)=\infty$, call $(X,\mu)$ UV-divergent if for any $\epsilon$ there is a measurable subset $A\subseteq X$ with $0<\mu(A)<\epsilon$.

Clearly $\Bbb N$ with the counting measure is IR-divergent but not UV-divergent while $[a,b]$ with the Lebesgue measure is UV-divergent but not IR-divergent.

Lemma

If $(X,\mu)$ is not IR-divergent then for any $p≤q\in[1,\infty)$ you've got $L^p(X,\mu)\subseteq L^q(X,\mu)$.

Proof:

Let $f\in L^q(X,\mu)$. Then
$$M(x):= \max(1, |f(x)|^q)$$
is integrable since $|f(x)|^q$ is integrable and since the constant function $1$ is integrable. Further $M(x)$ clearly majorises $|f|^p$ since $q≥p$ - hence $|f|^p$ is integable giving $L^p(X,\mu)\subseteq L^q(X,\mu)$.

Lemma

If $(X,\mu)$ is not UV-divergent then for any $p≤q\in[1,\infty)$ you've got $L^p(X,\mu)\supseteq L^q(X,\mu)$.

Proof:

If $f\in L^p(x)$ then $A:=|f(x)|^{-1}([1,\infty))$ has finite measure so by $X$ not being UV-divergent there exists a finite covering $A= \bigcup_{i=1}^n A_i$ where none of the $A_i$ admit sub-sets of smaller (non-zero) measure. In particular $|f(x)|$ must be constant almost everywhere on the individual $A_i$ as such $|f(x)|$ must admit an essential supremum on $A$, call this superemum $S$.

$$M(x):=\begin{cases} S^q & x\in A\\ |f(x)|^p& x\notin A\end{cases}$$
is integrable since $A$ has finite measure and $|f(x)|^p$ is integrable. Further it majorises $|f(x)|^q$ since outside of $A$ you have $|f(x)|<1$ whence $|f(x)|^q ≤ |f(x)|^p$ and inside of $A$ you have $|f(x)|^q ≤ S^q$. It follows that $L^p(X,\mu)\supseteq L^q(X,\mu)$.

So the thing you have seen is an expression of the fact that $\Bbb N$ does not admit subsets of arbitrarily small measure and that $[a,b]$ has bounded Lebesgue measure.


问:
Order of an operator given by a Fourier multiplier
I was given a Fourier multiplier $m(\xi)=|\xi|^{-\alpha}$ for $\alpha > 1$, and the operator $L$ associated with $m$ is given by: $$\widehat{Lf}(\xi) = m(\xi) \hat f(\xi).$$
How can we define the order of the operator $L$?  Is it different from the order of the multiplier $m$? I received mixed answers: $\alpha$ and $-\alpha$.

I am quite confused. Is there a reference for these definitions?

答:
With order do you mean the least $k$ such that the operator is bounded in the $H^k$ norm?

Your operator is unbounded in the $H^k(\Bbb{R})$ norm for all $k$: that $(1+|\xi|^k) \hat{f}(\xi)\in L^2(\Bbb{R})$ doesn't imply that $(1+|\xi|^k)|\xi|^{-\alpha} \hat{f}(\xi)\in L^2(\Bbb{R})$.


问:
What is the simplest function yielding Fibonacci numbers upon iteration?
A while ago I asked a question on trigonometric functions that are iteratively periodic. That is, after a finite number of iterations (compositions of the function with itself) the function returns to its original form. In other words, for some $k \in \mathbb{Z}_{>1}$ we have $f^{[k]} (x) = f^{[1]}(x) =: f(x)$.

In the question, I also mentioned some counterexamples that are not iteratively periodic. In these instances, I found functions that yield Fibonacci numbers upon iteration:

Consider $m(x) := m^{[1]}(x) = \cot(\arcsin(x)) = \frac{\sqrt{1-x^{2}}}{x}$. Then, $m^{[n]}(x) = \sqrt{ \frac{F_{n} - F_{n+1}x^{2}}{F_{n}x^{2} - F_{n-1}} \cdot (-1)^{n+1} } $ for $n >1$, where $F_{n}$ is the $n$'th Fibonacci number.

Define $l(x) := l^{[1]}(x) = \cos(\arctan(x)) = \frac{1}{\sqrt{1+x^{2}}} $. In this case, $l^{[n]} (x) = \sqrt{\frac{F_{n} + F_{n-1}x^{2}}{F_{n+1} + F_{n}x^{2}}}$ for $n>1$.

Question: though $m(\cdot)$ and $l(\cdot)$ are not the most complex functions, I wonder whether there exist functions that are even simpler than these and generate fibonacci numbers as well upon iteration. On could consider a rigorous definition of function complexity, but if that's not your thing then a simple function from a more intuitive sense (as little operations and inputs as possible) suffices as well. The function does not have to be trigonometric.

答:
Take $f(x) = 1+\frac{1}{x}$. Then $$f^{[n]}(x) = \frac{F_{n+1} x +F_n}{F_n x + F_{n-1}}$$

A related one is $g(x) = \frac{1}{1+x}$. Then $$g^{[n]}(x)=\frac{F_{n-1}x+F_n}{F_n x+F_{n+1}}$$

I doubt there are "simpler" functions that those two whose iterations involve Fibonacci numbers.

This functions also have other amusing properties, for example:

If you define $f^{[n]}(x)$ for $n<0$ as the iterations of the inverse function, and extend in the usual way the Fibonacci numbers to negative indices, the formula above remains valid.

We have $\displaystyle \lim_{n\to\infty} f^{[n]}(1) = \lim_{n\to\infty}\frac{F_{n+2}}{F_{n+1}}=\phi$, the golden ratio.


问:
Get rid of mixed derivatives in stationary PDE using coordinate transform
In this question the PDE
$$u_{xx} + u_{yy} + u_{zz} + u_{zy} = 0,$$
was rewritten to the standard Laplace equation $\Delta u=0$ using a coordinate transform, in some sense similar how you can map an ellipse to a circle and vice versa. I was wondering if a similar technique can be applied to the PDE
$$-u_{xx} - u_{xy} - u_{yy}  + 5u = f.$$
Possibly with Neumann boundary conditions.

答:
You can make change of variables: $\xi = y - \frac{x}2$, $\eta = \frac{x\sqrt{3}}2$, $v(\xi, \eta) = u(x, y)$. And you will get an equation
$$
5 v - \frac{3}4 \partial_\xi^2 v- \frac{3}4 \partial_\eta^2v = f,
$$
or
$$
\left(\Delta - \frac{20}3\right)v(\xi, \eta)=-\frac{4}3f(\xi, \eta).
$$
It is inhomogeneous Helmholtz equation. Such equations are studied separately from the Poisson and Laplace equations, but in a similar way.


问:
Prove that $2^{n} s$ can be written as the sum of perfect squares, where $s$ is the sum of perfect squares
I need require assistance in proving that $$2^{n} s$$ can be written as the sum of two perfect squares, where $$s$$ is a sum of two perfect squares. My teacher has told me I'm not allowed to use induction and I'm not really sure how to go about it.

If anyone could nudge me in the night direction, I'd be appreciative.

答:
In the case that $ n $ is odd, let $ s = k^2 $ and $ n = 2m + 1$. Hence $$ 2^{2m + 1}k^2 = \left(2^m k\right)^2 + \left(2^mk\right)^2 $$ Otherwise, $ n = 2m $ and we have that $$ 2^{2m}k^2 = \left(2^m k \right)^2 + 0^2 $$
No induction needed here. $ m $ and $ k $ are both in $ \mathbb{N} \cup \{0\} $.

You might think the zero is "cheating" but it is definitely necessary in some cases such as $ 2^n s = 16 $.


问:
Tricky Polar Line integral problem
I need to calculate the following line integral over the curve C:

$$\int_C e^{\sqrt{x^2+y^2}} ds $$

where $C$ is the circuit bounded by $r = a, \varphi = 0$ and $\varphi = \pi/4$.
I tried separating the integral into these three parts and got an answer of $(\pi/4)ae^a$
However, the answer the book gives (the problem is from Demidovich's Problems in Analysis) is

$$(\pi/4)ae^a + 2(e^a - 1)$$

The $e^a - 1$ term comes from evaluating the integral over the bounds of the curve $\varphi = 0$ and $\varphi= \pi/4$. But I thought these would cancel out, and not add up.

答:
a) Between point $(0,0)$ and $(a,0)$, $\vec r(t) = (t,0), 0 \leq t \leq a$.

$I_1 = \displaystyle \int_C f(\vec r(t)) \ |r'(t)| \ dt = \int_0^a e^t \ dt = e^a - 1$

b) Between point $(\frac a {\sqrt2},\frac a {\sqrt2})$ and $(0,0)$, $\vec r(t) = (\frac a {\sqrt2} - t, \frac a {\sqrt2} - t), 0 \leq t \leq \frac a {\sqrt2}$.

$I_2 = \displaystyle \int_0^{a / \sqrt2} \sqrt2 \ e^{(a - \sqrt2 t)} \ dt = e^a - 1$

c) over circular segment, as you calculated.

Title: Tricky Polar Line integral problem


答:
It's just $z^2y''+zy'-y=0$ where $z=x+1$. This is Cauchy-Euler's differential equation.
$$z^2y''+2zy'-zy'-y=0$$
$$(z^2y')'-(zy)'=0$$
$$(z^2y'-zy)'=0$$
Integrate.

Title: How to solve these type of differential equations: $y''+1/(1+x)y'-(1/(1+x)^2)y=0$

Body: Is there any particular way one can go about solving such questions. Here, I assumed $P=1/(x+1)$, $Q=-1/(x+1)^2$ and tried solving it but to no avail. Thanks in advance.