问:
Solving $\phi(n)=84$
Ok, I really need some help understanding this because either my brain isn't working at the moment or I'm breaking math and I have a striking suspicion that one of those is more likely.

Anyways, here's my process. WolframAlpha tells me there are 12 integer solutions, they are all $^+_-n$ so let's say there are 6 of them. They are as follows: $129, 147, 172, 196, 258, 294$

Ok, so here's how I attempted to solve it. $n$ looks like this ${\prod_{i=1}^kp_i^{\alpha_i}}$. And $\phi(n)=n\prod_{p\vert n}(1-\frac{1}{p})$. So I plug $n$ in and get an equation like this:

$\prod_{i=1}^k p_i^{\alpha_i-1}(p_i-1)=84$

Now, solving for $n$ SHOULD be equivalent to solving for $p_i, \alpha_i$, right?

So, I started going through possible values of $k$ and I'm interested in the case of $k=2$.

This means that $n$ has two prime factors, so it looks like $p_1^{\alpha_1}p_2^{\alpha_2}$, and the equation is:

$p_1^{\alpha_1-1}(p_1-1)p_2^{\alpha_2-1}(p_2-1)=84$

Ok. Here's my issue. Factoring $84$, we get $2\times2\times3\times7$

So I try to see a pattern. The first solution I came up with is $p_1=7, \alpha_1=2, p_2=2, \alpha_2=2$. Then I followed the same logic that led me to that solution but didn't find any more so I concluded that the other solutions must be as higher values of $k$. But then I factored $172$ and saw that it also had only two primes. It also fits the pattern. So I realized that I could get ones in that product as a result of $p^0$ and following that pattern, filling in the remaining factors, I found the $129$ solution.

No more, the next one must be at $k=3$. NOPE. $147$ also factors into only 2 primes. So now I'm really at a miss. Is there a more effective way to iterate over all of these combinations?

答:
Here is a technique you can use to solve these kinds of problems that doesn't need much intuition. Firstly, you need a lemma.

Lemma: If $n$ is odd, then $\phi(n) =\phi(2n)$.

If the prime factorisation of $n$ is $n = \prod_{i=1}^m p_i^{\alpha_i}$, then $\phi(n) = \prod_{i=1}^m p_i^{\alpha_i-1}(p_i-1)$. So $p_i - 1 \mid 84$. Listing out factors of 84, the possible values $p_i-1$ can take are:

$$p_i-1 = 1,2,4,6,7,12,14,21,42,84.$$

Then,
$$p_i = 2,3,5,7,8,13,15,22,43,85,$$ of which $8,15,22,85$ are not prime. So possible values of $p_i$ can take are:
$$p_i = 2,3,5,7,13,43.$$

Now condition on the highest prime occurring in the prime factorisation of $n$.

$43$ is the highest prime dividing $n$: Then $\phi(n) = 42*a, a=2$. Then we only need to hunt for something to contribute a factor of $2$ for $\phi(n)$. This can come from $2^2$ or $3$. This gives $n = 43*3$ or $n=43*2^2$ respectively. With the lemma, $\phi(2*43*3)=\phi(3*43)$. This gives $3$ solutions, $n= 129,258,172$.

$13$ is the highest prime dividing $n$. $\phi(n) = 12 *a , a=7$. We need to hunt for a term contributing $7$. $7$ can either come from one of the $p_i-1$'s or $p_i$'s. It is the latter because $p_i-1 = 7 \implies p_i = 8$, which is not prime. Yet, having $7 \mid n$ would imply $6 *12 \mid \phi(n)$, contradiction. So $13$ cannot be the highest prime dividing $n$.

$7$ is the highest prime dividing $n$. $\phi(n) = 6*a, a=14$. We need to hunt for a term contributing $14$. Yet all primes dividing $n$ are going to be smaller than $7$ already, so the only option for $\phi(n)$ to incur a factor of $7$ would be for the power of $7$ in the prime factorisation of $n$ to be $\geq 2$. We can further see that it has to be exactly $2$. This means that $7^2 \mid n$ and $\phi(n) = 7*6*a, a=2$. We still need a factor of $2$, which would come from $3$ or $2^2$. So $n = 7^2 * 3$ or $n = 7^2 * 2^2$. From the lemma, $n = 2*7^2 *3$ is a viable option as well. So $n = 147,294, 196$.

$5$ is the highest prime dividing $n$. Then we need to hunt for a term contributing $21$. Yet this implies $7 \mid n$, but now the primes are too `small' to contribute a $7$, contradiction.

Similarly, for $3$ and $2$ being the highest prime dividing $n$, it will always imply that $7 \mid n$. So there are no more answers.

Title: Solving $\phi(n)=84$


问:
Can two polynomials $p(x)$ and $q(x)\in \mathbb{F}[x]$ have just one factor in common?
Let $p(x),q(x)\in \mathbb{F}[x]$. EDIT: $\mathbb{F}$ is a field of $0$ characteristic.

Let us suppose there is an element $b\notin \mathbb{F}$ such that $p(b)=q(b)=0$. Then, $p(x)$ and $q(x)$ both have the factor $(x-b)$ in $\mathbb{F}[b,x]$.

My question is can $p(x)$ and $q(x)$, both of which belong to $\mathbb{F}[x]$, have only one common factor $(x-b)$ in $\mathbb{F}[b,x]$ and no other?

Motivation behind the question: Let me take you step by step.

If $f(x)\in \mathbb{F}[x]$, and $f(x)=g(x).h(x)$, where if $g(x)\in \mathbb{F}[x]$, then $h(x)$ also has to $\in \mathbb{F}[x]$. This is because if $h(x)$ contains coefficients from outside of $\mathbb{F}$, then, considering the coefficients of $g(x)$ are from $\mathbb{F}$, the coefficients of $g(x).h(x)$ wouldn't be from $\mathbb{F}$.

Let $f(x)\in \mathbb{F}[x]$ such that it has a root $b\notin\mathbb{F}$. Then, $f(x)=g(x).h(x)$ such that $g(x),h(x)\in \mathbb{F}[x]$, and $g(b)=0$. Obviously, $deg(g(x))\geq 2$, as if it is $1$, then it will be the linear factor $(x-b)$, which is not there in $\mathbb{F}[x]$.

Let us factorise $g(x)$ from point $(2)$ in $\mathbb{F}[b,x]$. $(x-b)$ is one factor. Let the other factor be $q(x)$. Then $q(x)\in \mathbb{F}[b,x]$, but $q(x)\notin \mathbb{F}[x]$ (from point $(2)$).

Now let us come back to the original question. If $p(x),q(x)\in \mathbb{F}[x]$ have a common root $b\notin F$, then $p(x)$ will factorise in the form $(x-b)t(x)f(x)$, where $t(x)\in \mathbb{F}[b,x]$ such that $(x-b).t(x)\in\mathbb{F}[x]$, and $q(x)$ will factorise in the form $(x-b)s(x)g(x)$ such that $s(x)\in\mathbb{F}[b,x]$ such that $(x-b)s(x)\in\mathbb{F}[x]$. It is clear that $t(x), s(x)\notin \mathbb{F}[x]$. Obviously$f(x),g(x)\in\mathbb{F}[x]$. $t(x)\neq s(x)$, as otherwise $p(x)$ and $q(x)$ won't have just $(x-b)$ in common. Also $f(x)\neq g(x)$.

Clearly, carrying forward from point $(4)$, $(x-b)t(x)$ and $(x-b)s(x)$ will form distinct irreducible polynomials in $\mathbb{F}[x]$, with no common factors. Hence, as $f(x)\neq g(x)$, $p(x)$ and $q(x)$ will not have any common factors in $\mathbb{F}[x]$, but will have the common factor $(x-b)$ in $\mathbb{F}[b,x]$. Isn't this a contradiction of the theorem that if two polynomials have a common factor in a field extension $\mathbb{K}$, then they have to have a common factor in the original field $F$, where  $\mathbb{F}\subset\mathbb{K}$?

The only way out of this contradiction that I can see is if $(x-b)t(x)$ can be factored into $r(x)e(x)$, and $(x-b)s(x)$ into $r(x)u(x)$, where $r(x), e(x)$ and $u(x)\in \mathbb{F}[x]$. $p(x)$ and $q(x)$ will then have the common factor $r(x)$ in $\mathbb{F}[x]$. I want to know if this is always possible.

答:
It's possible that the only linear common factor of $p$ and $q$ is $x-b$. When? Since the minimum polynomial of $b$ divides $p$ and $q$, it must be of the form $(x-b)^k$, for some $k$.

This can actually happen: consider the field $K=\mathbb{F}_2(t)$ of rational functions over the field $\mathbb{F}_2$ with two elements. Then the polynomial $x^2-t\in K[x]$ is irreducible and it's the minimum polynomial over $K$ of its unique root.

On the contrary, if the field has characteristic $0$, this can't happen: an irreducible polynomial $f$ (like the minimum polynomial for $b$) has distinct roots, because $f$ and its derived polynomial cannot have common non constant factors.

ContentLicense: CC BY-SA 3.0


问:
This series is known? $\sum _{n=1}^{\infty }\frac{2^{n-1}x^{2n-1}}{\prod _{i=1}^n\left(2i-1\right)}$
This series is known?
I was solving a differential equation by power series and my solution involves that series

$$\sum _{n=1}^{\infty }\frac{2^{n-1}x^{2n-1}}{\prod _{i=1}^n\left(2i-1\right)}$$

答:
Note that
$$
\prod _{i=1}^n\left(2i-1\right)=(2n-1)!!\tag1
$$
Therefore, the sum you are looking for is
$$
\sum_{n=1}^\infty\frac{2^{n-1}x^{2n-1}}{(2n-1)!!}\tag2
$$
For $u\ge0$, let
$$
f(u)=\sum_{n=1}^\infty\frac{(2u)^{n-\frac12}}{(2n-1)!!}\tag3
$$
Then
$$
\begin{align}
f'(u)
&=\frac{\mathrm{d}}{\mathrm{d}u}\sum_{n=1}^\infty\frac{(2u)^{n-\frac12}}{(2n-1)!!}\tag{4a}\\
&=\sum_{n=1}^\infty\frac{(2u)^{n-\frac32}}{(2n-3)!!}\tag{4b}\\
&=\frac1{\sqrt{2u}}+\sum_{n=1}^\infty\frac{(2u)^{n-\frac12}}{(2n-1)!!}\tag{4c}\\[3pt]
&=\frac1{\sqrt{2u}}+f(u)\tag{4d}
\end{align}
$$
Explanation:

$\text{(4a)}$: differentiate $(3)$

$\text{(4b)}$: differentiate term by term

$\text{(4c)}$: pull the $n=1$ term out front and re-index

$\text{(4d)}$: apply $(3)$

Therefore, rearranging $(4)$ and multiplying by the integrating factor of $e^{-u}$, we get
$$
\left(e^{-u}f(u)\right)'=e^{-u}f'(u)-e^{-u}f(u)=\frac{e^{-u}}{\sqrt{2u}}\tag5
$$
Integrating, noting that $f(0)=0$, yields
$$
\begin{align}
e^{-u}f(u)
&=\int_0^u\frac{e^{-t}}{\sqrt{2t}}\,\mathrm{d}t\tag{6a}\\
&=\sqrt2\int_0^{\sqrt{u}}e^{-t^2}\,\mathrm{d}t\tag{6b}\\[6pt]
&=\sqrt{\pi/2}\operatorname{erf}\left(\sqrt{u}\right)\tag{6c}
\end{align}
$$
Explanation:

$\text{(6a)}$: integrate $(5)$

$\text{(6b)}$: substitute $t\mapsto t^2$

$\text{(6c)}$: apply the definition of $\operatorname{erf}$

Thus, $(6)$ yields
$$
f(u)=e^u\sqrt{\pi/2}\operatorname{erf}\left(\sqrt{u}\right)\tag7
$$
Therefore, we can apply $(3)$ and $(7)$ to get
$$
\begin{align}
\sum_{k=1}^\infty\frac{2^{n-1}x^{2n-1}}{(2n-1)!!}
&=\frac1{\sqrt2}f\!\left(x^2\right)\tag{8a}\\
&=\frac{\sqrt\pi}2e^{x^2}\operatorname{erf}(x)\tag{8b}
\end{align}
$$
Since the series and $\operatorname{erf}(x)$ are odd functions, we can remove the positivity restriction.

ContentLicense: CC BY-SA 4.0


问:
Exercise 9.8.5 in Analysis 1 by Tao (continuous function)

(a) For given $x, y \in \mathbb{R}$ with $x < y$, we know from proposition 5.4.14 that there exists $r' \in \mathbb{Q}$ such that $x < r' < y$. Since $q$ is bijection, there exists $n'$ such that $q(n') = r'$. Therefore, $f(x) < f(x) + g(q(n')) = f(x) + 2^{-n'} \le f(y)$.

(b) Note that $x > q(n') = r'$. Therefore, $f(x) = \sum_{r \in Q; r<x} g(r) \ge \sum_{r \in Q; r<r'} + 2^{-n'} = f(r')+2^{-n'}$. Consider the sequence $x_m = r' + \frac1n$. $(x_n)_{m=1}^\infty$ converges to $r'$ and every $x_m > r'$. Thus,$\lim_{m\to\infty}f(x_m) \ge f(r')+2^{-n'} >f(r')$. Thus, $f(x)$ is not continuous at rational numbers.

I am struggling with (c). I appreciate if you give some help.

答:
Hint/Sketch:

Fix an irrational $x \in \mathbb{R}\setminus \mathbb{Q}$ and let $\varepsilon > 0$.  From here, we may define $N = \max(0,\lceil-\log_2\varepsilon\rceil)$ and $$\delta = \frac{1}{2}\min_{0\leq n \leq N}(|q(n)-x|).$$  Show that $0 < \delta < \infty$ and if $|q(n)-x| \leq \delta$, then $n > N$.

In particular, this means
$$f(x+\delta) - f(x-\delta) = \sum_{\substack{r \in \mathbb{Q} \\ x-\delta \leq r < x+\delta}}g(r) \leq \sum_{n>N} 2^{-n} = 2^{-N} < \varepsilon.$$

To finish, suppose $y\in \mathbb{R}$ such that $|x-y| < \delta.$  Then since $f$ is monotone increasing, $$|f(x)-f(y)| \leq f(x+\delta) - f(x-\delta) < \varepsilon.$$


问:
Details on matrix multiplication
For Ax = b,

b is a linear combination of the columns of A. A scales x.

However, for AB, assuming both have suitable sizes, which matrix acts on which? Is there a linear combination of some sort going on? How do the columns/rows interact with each other? For example, 2x2 matrices.

答:
Giving a linear space $V$ and $W$ over a field $F$ (for example $V$ can be the vector space $\mathbb R^n$ where $F=\mathbb R$), a linear map is a function $f: V \to W$ such that for all $\mathbf v, \mathbf w \in V$ and $a \in F$, $$f(a\mathbf v+\mathbf w)=af(\mathbf v)+f(\mathbf w).$$

A matrix $\mathbf A$ is a linear map that takes a vector "$\mathbf v$" and returns a vector "$\mathbf{Av}$". In fact all linear functions can be defined like that, so we don't distinguish a matrix and a linear function.

Matrix multiplication is just function composition. It is defined as $\mathbf{AB} = \mathbf{A\circ B}$. In other words, it is defined so that $\mathbf{A(Bv)}=\mathbf{(AB)v}$ for all $\mathbf{v}$.

To prove that all linear map can be written as a matrix, we can write any vector $\mathbf{v} \in V$ as a linear combination of the basis $\{\mathbf{e_1,e_2,e_3,...e_n}\}$ of $V$:
$$
\mathbf{v}=v^1\mathbf{e_1}+v^2\mathbf{e_2} ... v^n\mathbf{e_n}
$$

(the superscripts are indices, not powers)

(A basis is a set of $n$ non-zero vectors in a $n$ dimensional vector space that are linearly independent, i.e. cannot be written as a linear combination of other vectors in the set.)

At the same time, we can define $\{\mathbf{e'_1, e'_2,...e'_m}\}$ to be a basis of $W$.

A linear map $f: V \to W$ applied by $v$ can be written as
$$
f(\mathbf v)= v^1 f(\mathbf {e_1}) + v^2 f(\mathbf {e_2}) + ... + v^n f(\mathbf {e_n}).
$$

But we can write
$$
f(\mathbf{e}_k)=w^1{}_k\mathbf{e'_1}+w^2{}_k\mathbf{e'_2}+...+w^m{}_k\mathbf{e'_m}
$$
therefore
$$
f(\mathbf{v}) = \sum_k\sum_j w^j{}_k v^k \mathbf{e'_j}
$$
If we see $\{v^k\}$ as a representation of $\mathbf v$, and the $\{w^j{}_k\}$, a table of $m\times n$ numbers, is a complete description of $f$.

Given $f: U\to V$ and $g: V \to W$, and
$$
f(\mathbf e_i)= \sum_j a^j{}_i \mathbf e'_j
$$
$$
g(\mathbf e'_j)= \sum_k b^k{}_j \mathbf e''_k
$$
then
$$\begin{align*}
g(f(\mathbf e_i)) &= g\left(\sum_j a^j{}_i \mathbf e'_j\right) \\
&= \sum_j a^j{}_i g(\mathbf e'_j) \\
&= \sum_j a^j{}_i \sum_k b^k{}_j \mathbf e''_k \\
&= \sum_k \sum_j a^j{}_i b^k{}_j \mathbf e''_k \\
\end{align*}$$

Therefore we can use $\left\{w^k{}_i=\sum_ja^j{}_i b^k{}_j\right\}$ to represent the linear mapping $g\circ f$.

Title: Details on matrix multiplication


问:
How to prove that UFD implies normal?
How to prove that UFD implies normal? or is there is a reference containing this proof?

a ring $R$ is a normal ring if $R$ is an integral domain that is integrally closed in $\operatorname{frac}(R).$

And $R$ is integrally closed in a commutative ring $S$ iff the integral closure of $R$ in $S$ (int($R$)) equals $R$. And int($R$) consists of all $x \in S$ that is integral over R. And note that an element is integral over $R$ if there exists a monic polynomial $f(x) \in R[x]$ such that $f(x) = 0.$

答:
The proof is the same as in the case of integers and rationals.

Let $R$ be a UFD and $Q$ its field of fractions. Let $q = \frac{a}{b} \in Q$ be such that $q$ satisfies a monic polynomial over $R$.

WLOG, we may assume that $a$ and $b$ have no common prime factors. We show that $b$ is a unit and hence, $q \in R$.

Let $f(x) = x^n + a_1x^{n-1} + \cdots + a_n \in R[x]$ be a polynomial that $q$ satisfies. In other words, $f(q) = 0$. Multiplying this equation with $b^n$ gives
$$a^n + a_1ba^{n-1} + \cdots + a_nb^n = 0$$
or $$a^n = -b(\underbrace{\cdots}_{\in R}).$$
Thus, every prime factor of $b$ is also one of $a$. By hypothesis, there is no such prime. Thus, $b$ must be a unit.

Title: How to prove that UFD implies normal?


问:
Area of intersection between two circles
Suppose you have 2 circles that intersect each other in such a way that each circle passes through the other's center. What is the area between the circle(or common area) i.e. area between the centres of the circles?

答:
Label the center of the first circle $C$ and the center of the second circle $C'$. Label one of the points of intersection of the two circles $A$ and the other $B$. Let the radius of the circles be $r>0$. It should be clear that the following lengths are all equal to $r$. $AC$, $AC'$, $BC$, $BC'$, $CC'$. With a simple application of Pythagoras' Theorem, we get that the length of the line segment $AB$ is $\sqrt{3}r$.

With some basic trigonometry, we find the angles $\angle ACB=\angle AC'B=\dfrac{2\pi}{3}$. So, the area of one half of the intersection is the area of a circular segment with angle $\theta=\dfrac{2\pi}{3}$ and radius $r$, which gives an area of $\dfrac{r^2}{2}(\theta-\sin\theta)=\dfrac{r^2}{2}\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right)$ and so the area of the entire intersection is twice this. This gives an area of $$r^2\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right).$$


问:
Calculate the residue of $f(z) = \frac {e^{-z}} {(z+2)^3} $
I'm currently studying integral solving methods using the Residue Theorem. However, I sometimes have problems finding a way to calculate the residue(s). For example, what do I do when a function includes the exponential function (or any other function whose Taylor series is commonly known)?

After a while, I found a solution that I added below this question. In general, I thought that it might be helpful to others when they are dealing with solving integrals using the residue theorem for functions that involve the exponential function (or any variants of it).

My initial problem was:

How do I calculate the residue of $f(z) = \frac {e^{-z}} {(z+2)^3} $ at $z = -2 ?$

With the solution to this problem, I was then able to calculate the integral using the residue theorem:
$$
\int_{|z|=3} f(z) dz =  ?
$$

答:
Since I'm fairly new to residue calculation, I had a lot of trouble with this one. After a while, I found a neat solution  that I'd like to share.
We have
$$\begin{align}
f(z) &= \frac {e^{-z}} {(z+2)^3} = \frac{1}{(z+2)^3} \sum_{k=0}^\infty \frac{(-z)^k}{k!} \\
&= \frac{1}{(z+2)^3} \sum_{k=0}^\infty \frac{(-1)^k}{k!}(z+2)^k e^2,
\end{align}$$
since
$
\sum_{k=0}^\infty \frac{(-1)^k}{k!}(z+2)^k = e^{-z-2}.
$
Thus we get
$$
f(z) = \frac{1}{(z+2)^3} \sum_{k=0}^\infty \frac{(-1)^k}{k!}(z+2)^k e^2 = e^2 \sum_{k=0}^\infty \frac{(-1)^k}{k!}(z+2)^{k-3}
$$
and the residue would be $\frac{e^2}{2}.$


问:
Probability Measure Preserving (pmp) Isomorphism and Automorphism
See section 1.4 of enter link description here. Here is an excerpt from that section:

...two probability measure spaces $(X_1, \mu_1)$ and $(X_2,\mu_2)$ are isomorphic if there exist conull subsets $Y_1$ and $Y_2$ of $X_1$ and $X_2$, respectively, and a Borel isomorphism $\theta : Y_1 \to Y_2$ such that $\theta_{*} \mu_1 |_{Y_1} = \mu_2 |_{Y_2}$, i.e., $(\theta_{*} \mu_1 |_{Y_1})(E) = \mu_1(\theta^{-1}(E)) = \mu_2(E)$ for every Borel subset $E$ of $Y_2$. Such a map $\theta$ is called a probability measure preserving (p.m.p.) isomorphism, and a p.m.p. automorphism whenever $(X_1, \mu_1) = (X_2,\mu_2)$...We denote $\text{Aut}(X,\mu)$ the group of (classes modulo null sets of ) p.m.p. automorphisms of a probability measure space $(X,\mu)$.

First, what exactly is a Borel isomorphism? Next, I'm guessing that $(\theta_{*} \mu_1 |_{Y_1})(E) = \mu_1(\theta^{-1}(E))$ is a definition. Is that right? Finally, I having trouble making sense of the definition of $\text{Aut}(X,\mu)$. Presumably, the group operation is function composition, but how does composition make sense when these p.m.p. automorphisms are defined on subsets of $X$ (conull subsets to be exact)? The domains ranges might not be compatible, no?

答:
A Borel isomorphism is a measurable bijection with measurable inverse. (Usually this term is only used for standard Borel spaces, I'm not sure if there is such an assumption here).

Yes, that's the definition of pushforward measure.

Elements of $\mathrm{Aut}(X,\mu)$ are equivalence classes of functions that agree modulo a null set. If $f$ is defined on $Y_1$ and $g$ is defined on $Y_2$ you can take $g\circ f$ to be defined on $f(Y_1)\cap Y_2$ (this makes sense because $f(Y_1)$ is still full measure).


问:
Whether Hausdorff Property is Required

Let $H$ denote a group that is also a Hausdorff topological space. Show that $H$ is a topological group if and only if the map $h: H \times H \rightarrow H, h:=(x,y)\rightarrow xy^{−1}$ is continuous.

The following proof does not use the Hausdorff property.

Suppose $h$ is continuous.

$f:H \rightarrow H \times H, f(x):=(e, x)$ is continuous (Munkres theorem 19.6). $(h\circ f)(x)=x^{−1}$ is continuous.

$j:H\times H \rightarrow H \times H, j(x, y):=(x, y^{−1})$. $\pi _1 (x, y)=x$ is continuous. $(h\circ f \circ \pi _2)(x, y)=y^{−1}$ is continuous. By Munkres theorem 19.6, $j$ is continuous. $(h\circ j)(x, y)=xy$ is continuous.

So $H$ is a topological group.

Suppose $H$ is a topological group. $k(x, y):=xy$ and $l(x):=x^{−1}$.

$\pi _1 (x, y)=x$ is continuous. $(l \circ \pi _2) (x, y)=x^{−1}$ is continuous. By Munkres theorem 19.6, $j$ is continuous. $j \circ k=h$ is continuous.

Please let me know whether Hausdorff property is required. (I'm aware that a topological group is Hausdorff).

Thanks.

Edit: Thanks to Aryaman Maithani's comment, some textbooks (for eg Munkres) require Hausdorff-ness in the  definition of topological group. Hausdorff-ness is required probably only in that case.

答:
Indeed the proof does not require any Hausdorffness, we only need that certain other natural product maps are continuous, and this holds regardless of separation axioms.

I would do it this way: suppose $h$ is continuous, then indeed $f:x \to (e,x)$ from $H$ into $H \times H$ is continuous (the composition with the two projections are a constant map (always continuous) resp the identity on $H$ (ditto). The inversion map $I:x \to x^{-1}: H \to H$ then is continuous as $h \circ f$, a composition of continuous maps. The product map $P: (x,y) \to xy$ is then also continuous as $h \circ (1_H \times I)$ etc.

And if $P$ and $I$ are continuous so is $h$ as the composition  $P \circ (1_H \times I)$, so both implications hold.

Title: Whether Hausdorff Property is Required


问:
Existence of a non-constant continuous function
Let $(X , d)$ be a connected metric space and let $(Y , d)$ be a discrete metric space. Then I know that every continuous function from $(X , d)$ to $(Y , d)$ is a constant function.

So is it true that there always exist a non-constant function from a connected metric space $(X , d)$ to a non-discrete metric space $(Y ,d)$. I am struggling to prove it. Please help. Thanks in advance.

答: Not true. Let $X$ be the real line with the usual metric and $Y=\{0,\frac 1 2, \frac 1  3,...\}$ with the usual metric. If $f$ is a non-constant continuous function from $X$ to $Y$ then IVP property is violated since $Y$ is countable. Note that $0$ is not an isolated point of $Y$ so the $Y$ is not discrete.


问:
What is the confidence interval for the mean of the population?
Let $X$~$Exp(\lambda)$ then
$f$($x$;$\lambda$) $=$ $\lambda$e$^{-\lambda x}$ $I$$_{(0,\infty)}$($x$). Obtain  confidence interval for the mean of the population.

To do this, we need to remember a few facts about the gamma distribution. More specifically, If $Y=X_1+X_2+\cdots+X_n$, where the $X_i$'s are independent $Exp(\lambda)$ random variables then $Y \sim Gamma(n, \theta)$. Thus, the random variable $Q$ is defined as $Q=\theta (X_1+X_2+\cdots+X_n)$ has a $ Q \sim Gamma(n, 1)$ distribution. Let us define $\gamma_{p,n}$ as follows. For any $p \in [0,1]$ and $n \in \mathbb{N}$ we define $\gamma_{p,n}$ as the real value for which $P(Q > \gamma_{p,n})=p,$ where $ Q \sim Gamma(n, 1)$ the Confidence interval is $\frac{\gamma_{1-\frac{\alpha}{2},n-1}}{X_1+X_2+\cdots+X_n}, \frac{\gamma_{\frac{\alpha}{2},n-1}}{X_1+X_2+\cdots+X_n}$. Is this right?

答:
Observe that

$$Y=2\lambda\Sigma_i X_i\sim \chi_{(2n)}^2$$

Now $Y$ is your pivotal qty so first calculate the 2 quantiles of $Y$ with paper tables or calculator and then you can easy find the confidence bounds for $\frac{1}{\lambda}$ solving the double inequality in $\frac{1}{\lambda}$.

Thus, setting $a,b$ the 2 quantiles of the $\chi_{(2n)}^2$ you get

$$a<2\lambda \Sigma_i X_i<b$$

That is

$$\frac{2\Sigma_i X_i}{b}<\frac{1}{\lambda}<\frac{2\Sigma_i X_i}{a}$$


问:
If $\alpha \preceq \beta$ then there is a unique ordinal $\gamma$ such that $\alpha + \gamma = \beta$
I tried to prove this proposition, if $\alpha \preceq \beta$ then there is a unique ordinal $\gamma$ such that $\alpha + \gamma = \beta$, by order type not by transfinite induction.

My attempt:

$W = \{ \xi \in \beta \mid \alpha \preceq \xi \}$ is a set of ordinals so, $W$ is a woset.

Every woset is order isomorphic to a unique ordinal, therefore, exist a unique ordinal $\gamma$ such that, $W$ and $\gamma$ are ordered isomorphic.

My problem:
I strongly believe that, $\gamma$ is the solution but I cannot prove that, $\alpha + \gamma = \beta$.

Def: $\prec := \in$. The ordinal addition is defined by induction.

I have read some similar questions. For example, (1) just talk about transfinite induction (limit case) and (2) just talk about the uniqueness. I hope your helps.

答:
I have proved one theorem:

If two disjoint ordered sets, $(W_1, \preceq_1)$ order isomorphic to $\alpha$ and $(W_2, \preceq_2)$ order isomorphic to $\gamma$, then ordered sum of the two ordered sets, $(W_1, \preceq_1) \oplus (W_2, \preceq_2)$, order isomorphic to $\alpha + \gamma$.

Def: $\forall x,y \in (W_1 \cup W_2, \preceq) = (W_1, \preceq_1) \oplus (W_2, \preceq_2): x \preceq y$ is defined:

$x,y \in W_1 \wedge x \preceq_1 y$ or

$x,y \in W_2 \wedge x \preceq_2 y$ or

$x \in W_1 \wedge y \in W_2$.

Back to the problem above, from $\alpha \cup W = \beta$ and $W$ order isomorphic to $\gamma$ so, $\alpha \oplus W = \beta$ order isomorphic to $\alpha + \gamma$ by the theorem. Therefore, $\alpha + \gamma = \beta$.

Just find out a way to explain.


问:
Does $(ab)^n=(ba)^n$ with $a^n=e$ imply $ab=ba$ in a group $G$?
Let G be a group and $a,b \in G$. Moreover, suppose $(ab)^n=(ba)^n=b$ for some $n \in \mathbb{Z}$. Here $a^n=e$ but $a$ is not necessarily the identity element. Can we conclude $ab=ba$?

I know in general it is not true that $(ab)^n=(ba)^n \Rightarrow ab = ba$. But in this case, $a^n=e$, which might change the situation. I cannot make the conclusion that the implication is entailed.

答:
Here's the reasoning:

$$(ab)^n=b$$
$$(ab)^na=ba$$
$$a(ba)^n=ba$$
$$ab=ba$$

I know in general it is not true that $(ab)^n=(ba)^n \Rightarrow ab = ba$. But in this case, $a^n=e$, which might change the situation. I cannot make the conclusion that the implication is entailed.

Note that $a^n=e$ is irrelevant. What really matters is that not only $(ab)^n=(ba)^n$ but more importantly that both are equal to $b$.


问:
Systematic approach to solve logical problem
I came across following problem and corresponding solution:

When I tried it myself I was not able to come up with the solution, especially the contrapositive step. May be because never in the past I tried rewriting given sentences in another (contrapositive) form and solve from there. So I was guessing can there be any non-tricky approach to solve such problems. My go to approach is to utilise resolution refutation. I came up with following clauses:

$\neg m\vee i \quad\text{(from P1)}$

$m \vee \neg i \quad \text{(from P2)}$

$m \vee l \quad \text{(from P2)}$

$\neg i \vee h \quad \text{(from P3)}$

$\neg l\vee h \quad \text{(from P3)}$

$\neg h\vee g \quad \text{(from P4)}$

($m$ for mythical, $i$ for immortal, $l$ for mammal, $h$ for horned, $g$ for magical)
To this I added $\neg m$, $\neg h$, $\neg g$ and tried to derive empty clause / $FALSE$. But I was not able to. Q1. Is this enough?

Q2. Is there any systematic / less tricky approach to solve such problems?

答:
The clauses you have written are ok.

Now, we have to use Resolution to derive that it is Horned and Magic.

For Horned we need the first three premises with the negation of the purported conclusion: $\lnot h$:

$¬m ∨ i$

$m ∨ ¬i$

$m ∨ l$

$¬i ∨ h$

$¬l ∨ h$

$¬h$ --- negated conclusion

$i \lor l$ --- from 1) and 3)

$l \lor h$ --- from 7) and 4)

$h$ --- from 8) and 5)

$\square$ --- from 6) and 9)

Now, having proved $h$ from P1)-P3), we can use P4) $\lnot h \lor g$ to prove Magic:

$h$

$\lnot h \lor g$

$\lnot g$ --- negated conclusion

$g$ --- from 1) and 2)

$\square$ --- from 3) and 4).

We cannot prove that it is Mythical. If we add $\lnot m$ to the set of clauses, we cannot produce the empty clause.


答: No.  Let $X$ be any connected metric space (for example, $[0,1]$) and $Y$ any totally disconnected metric space; that is, one in which every connected component is a single point (for example, $\mathbb{Q}$).  Then $C(X\to Y)$ contains no non-constant functions.  For: if $f\in C(X\to Y)$, then $\mathrm{rng}{(f)}$ is connected, and so contained in a connected component of $Y$, which must be a singleton.

Title: Existence of a non-constant continuous function

Body:
Let $(X , d)$ be a connected metric space and let $(Y , d)$ be a discrete metric space. Then I know that every continuous function from $(X , d)$ to $(Y , d)$ is a constant function.

So is it true that there always exist a non-constant function from a connected metric space $(X , d)$ to a non-discrete metric space $(Y ,d)$. I am struggling to prove it. Please help. Thanks in advance.


问:
Combinations: If n players in team, each has to play x out of y games what is the maximum number of players one can have for all matches?
I think this can be solved with a combinations formula. But, not quite sure which formula to use.

Combinations: If n players can play on a team and all players have to play at least x out of y (y>x) games, then what is the maximum number of players one can have for all the matches together?

Example: If 12 players can play on a team and all players have to play at least 3 out of 4 games then what is the maximum number of players one can have for all the matches together?

The answer would be 16.

What would be a math formula to generalize this?

答:
Converted from comment as requested:

If the squad of players is size $s$ then the proportion of matches they can play in on average is $\frac n s$.  This must be at least $\frac x y$ so $\frac n y \le \frac x y$, implying $s \le \frac{ny}{x}$ and since $s$ is an integer, you need to round down and its maximum possible value is $$\Big\lfloor\dfrac{ny}{x}\Big\rfloor$$


问:
Triangles in a cyclic quadrilateral
Consider a cyclic quadrilateral $ABCD$ where

$|DA|\lt |AB|=|BC|\lt |CD|$, $E:|BE|\bot |AC|, E\in |CD|$  , $F:|EF|\text{ }||\text{ }|BC|,F\in |AB|$.

The task is to prove that |FB|=|FD|, that is, that $\Delta FDB$ is isosceles.

I have been able to prove some statements about the quadrilateral, like tha fact that the triangles formed by the diagonals opposite each other have to be similar, but I am struggling to prove anything of use regarding $\Delta ABC$ and $\Delta FDB$.

Thank you for any help.

答:
Draw $AE$ and $FD$.

Observe that, $AFED$ is a cyclic quadrilateral.

Also, since $\triangle ABC$ is isosceles, $BE$ is the perpendicular bisector of $AC$.

Thus, $\angle FDB=\angle FDE-\angle BDE=\angle FDE-\angle BDC=\angle FAE-\angle BAC=\angle FAE-\angle FAC=\angle CAE=\angle ACE=\angle ACD=\angle ABD=\angle FBD$.

Therefore, $\triangle FDB$ is isosceles.

Edit : The reason why $AFED$ is cyclic is because since $EF$ is parallel to $BC$, $\angle AFE=\angle ABC=180-\angle ADC=180-\angle ADE$ and hence $AFED$ is cyclic.[It's opposite angles are supplementary.]   ( Credit : User @V.S.e.H.)

Title: Triangles in a cyclic quadrilateral


问:
How to prove $G_2$ is a normal subgroup of $G_1\times G_2$?
I am studying abstract algebra and in one of my previous midterms there was a question that I can not solve. Now I have final exam tomorrow and want to learn the answer to that question. Here it is:

Can anyone help me with the solution of this question? My answer is there, but it would be very nice if you assume there is no answer and explain the answer to the question for part (i) and (ii), I really want to understand it. Thank you in advance

答: Hint: Show that any element of $G_1$ commutes with any element of $G_2$.


问:
Prove that $ k[x,y]/(x^3 - y^2)$ is not a UFD.
I want to prove that: the integral domain $k[x,y]/ (x^3 - y^2)$ is not normal and find its integral closure in its field of fractions.

I know that I will use the idea that $UFD \implies$ normal and hence not normal means not $UFD$ so what I should show is that $k[x,y]/ (x^3 - y^2)$ is not a UFD and I have read a hint here How to show a ring is normal or not, and how to show the normalisation of the ring to consider $x = (y/x)^2.$but I can not see how that will show that it is not a UFD. could someone elaborate this to me please?

Also, what is the field of fraction of this integral domain?

I know that a unique factorization domain is an atomic integral domain in which factorization to irreducibles is unique (up to associates). And a commutative ring is atomic if each $r \in R$ is a finite product of irreducibles in R.

Also, a ring $R$ is a normal ring if $R$ is an integral domain that is integrally closed in $\operatorname{frac}(R).$

And $R$ is integrally closed in a commutative ring $S$ iff the integral closure of $R$ in $S$ (int($R$)) equals $R$. And int($R$) consists of all $x \in S$ that is integral over R. And note that an element is integral over $R$ if there exists a monic polynomial $f(x) \in R[x]$ such that $f(x) = 0.$

答:
Well, I think the first step is to find the standard basis of the quotient ring, which is

$\{x^iy^j\mid 0\leq i\leq 2, j\geq 0\}.$


问:
Proving $\lim \sup \sin n = 1$
I'm trying to prove that $\limsup \sin(n) = 1$. This fact sounds obvious to me, but I don't know how to use the given hint that any numbers of the form $n + 2m\pi$, $m,n \in \mathbb{Z}$, are dense in the real numbers. (I also can't figure out how to prove this, aside from taking $a < b$, defining the positive difference $b - a$, and using density of the irrationals.)

Any hints or help would be appreciated.

答: The superior limit of a sequence determines the limit of the largest convergent subsequence. Let $$S=\{\sin(n): n\in \mathbb{Z}\}.$$ We already know that that $\sin(x)$ is upper-bounded by $1$, and if we consider the sequence $$\tilde{S}=\{\sin(\pi n):n\in \mathbb{Z}\},$$ we can clearly see that $$\lim\sup \tilde{S}=1.$$ The main issue is that $$\{\pi n:n\in\mathbb{Z}\}\not\subset\mathbb{Z}\quad\forall n.$$ This is where the hint is needed. Although the points in $\{\pi n:n\in \mathbb{Z}\}$ do not belong in $\mathbb{Z}$ (except for $n=0$), thanks to $\{n+m\pi:n,m\in\mathbb{Z}\}$ being dense in $\mathbb{Z}$, given $n$, we can find $\tilde{n}, m\in \mathbb{Z}$ so that $\tilde{n}$ is as close as we want to $m+\pi n$ $\forall n$.

Title: Proving $\lim \sup \sin n = 1$


问:
Solving $A*X*B = X*C$ with $X$ the only unknown
I've formulated a problem as above, and wondering if a solution is actually possible.

Dimensionality:

$A$ is $N \times N$ (known)

$B$ and $C$ are $N \times 1$ (known)

$X$ is $N \times N$ (unknown)

Now obviously the problem is very underdetermined, and I am also expecting the solution would involve null space, but I am not worried about this for now as I can add an arbitrary number of observations.

But before I do that, I am just wondering if it is possible in principle to disentangle X from the sandwich it finds itself in?

答:
Here is a partial answer:

If $C=\lambda B$ then $XB$ is an eigenvector of $A$ of eigenvalue $\lambda$, you can compute all of them and then define any $X$ that sends $B$ to such eigenvector.

Otherwise they are linearly independent, so you can extend them to a base of the whole space. Let $v$ be any $N\times 1$ vector and $w=Av$. Then you can find $X$ such that $XB=v$ and $XC=w$ by defining a linear transformation on the basis we created before (lots of liberty here), any such $X$ is going to be a solution.


问:
Eigenvalues of integral operator
Let $f \in L^{p}[0,1]$ for $p \in (1,\infty)$ and consider the operator $Tf(u) = \int_{0}^{u} f(t) dt$. We want to show that $T$ has no eigenvalues. An eigenvalue $\lambda \in \mathbb{R}$ means that there is a non-trivial solution to the equation
$$
\lambda f(u) = \int_{0}^{u} f(t) dt
$$
for all $t \in [0,1]$.

Note that this is easy to do when $f$ is assumed to be continuous. We differentiate both sides to find the $f$ has to be something exponential and realize that $f(0) = 0$ (thus no solution). However, when $f$ is simply in $L^{p}$ but not continuous we cannot do this. My thought is to use the density of continuous functions in $L^{p}$ to do this, but I'm not able to formalize the argument.

My "argument" flows like this: If for fixed $\lambda \in \mathbb{R}$ has the property that
$$
\int_{0}^{u} f(t) dt - \lambda f(u) = 0
$$
for all $u \in [0,1]$, this implies that there is a continuous $g$ such that $\int_{0}^{u} g - \lambda g = 0$, which is a contradiction. However, I'm not sure how to prove the existence of such a $g$. Is it even true?

答:
Continuity of the operator: Note that
$$Tf(x)=\int_0^xf(t)dt=\int_0^1f(t)\cdot\chi_{[0,x]}(t)dt$$
so
$$\|Tf\|_p^p=\int_0^1\int_0^1|f(t)|^p\cdot\chi_{[0,x]}(t)dtdx=\int_0^1\bigg(|f(t)|^p\int_0^1\chi_{[t,1]}(x)dx\bigg)dt=$$
$$=\int_0^1|f(t)|^p\cdot(1-t)dt\leq\int_0^1|f(t)|^pdt=\|f\|_p^p$$
so $T$ is bounded and $\|T\|\leq1$. For the eigenvalues, let $\lambda\neq0$ be a scalar so that there exists a non-zero function $f\in L^p[0,1]$ such that  $Tf(x)=\lambda f(x)$, so we have that
$$f(x)=\frac{1}{\lambda}\int_0^xf(t)dt$$
Observe that
$$|f(x)-f(y)|=\frac{1}{|\lambda|}\bigg|\int_x^yf(t)dt\bigg|\leq\frac{1}{|\lambda|}\int_x^y|f(t)|dt=\frac{1}{|\lambda|}\int_0^1|f(t)|\chi_{[x,y]}(t)dt\leq$$
(using Holder's inequality)
$$\leq\frac{1}{|\lambda|}\|f\|_p\cdot\big(\int_0^1\chi_{[x,y]}(t)dt\big)^{1/q}=\frac{\|f\|_p}{|\lambda|}\cdot|x-y|^{1/q}$$
so $f$ is in $\text{Lip}(\frac{\|f\|_p}{|\lambda|},\frac{1}{q})$ and in particular it is continuous. But since $f(x)=\lambda^{-1}\int_0^xf(t)dt$, it is not only continuous, but differentiable too! Taking derivatives we see that $\lambda f'=f$. This has a standard solution, something with a constant multiple of the exponential, I cannot recall at the moment. I believe after plugging this standard solution in you will immediately get to the conclusion that $f=0$, a contradiction.

Title: Eigenvalues of integral operator


答:
Look at $\varphi : G_1 \times G_2 \rightarrow G_1, (g,h) \mapsto g$. Why is this a homomorphism? What is its kernel?

This is enough because the normal subgroups of $G_1 \times G_2$ are exactly the kernels of group homomorphisms out of $G_1 \times G_2$.

Title: How to prove $G_2$ is a normal subgroup of $G_1\times G_2$?

Body:
I am studying abstract algebra and in one of my previous midterms there was a question that I can not solve. Now I have final exam tomorrow and want to learn the answer to that question. Here it is:

Can anyone help me with the solution of this question? My answer is there, but it would be very nice if you assume there is no answer and explain the answer to the question for part (i) and (ii), I really want to understand it. Thank you in advance


问:
Given a sequence of sets $\{E_j\}$, define $F_1=E_1$ and $F_j=E_j-(E_1\cup\dots\cup E_{j-1})$ for $j>1$. Why does $\bigcup E_j=\bigcup F_j$?
$\{ E_j \}_{j=1}^{\infty}$ : sequence of sets

We define $\{ F_j \}_{j=1}^{\infty}$ with
\begin{align}
&F_1=E_1 \\
&F_j=E_j-(E_1 \cup \cdots \cup E_{j-1}) \ (j\geqq 2)
\end{align}

Then, prove that

・$ F_j \cap F_k= \varnothing$  for $j\neq k$.

・$\bigcup_{j=1}^{\infty} E_j =\bigcup_{j=1}^{\infty} F_j $.

I could prove $ F_j \cap F_k=\varnothing$  for $j\neq k$ but I cannot prove $\bigcup_{j=1}^{\infty} E_j =\bigcup_{j=1}^{\infty} F_j $.

Because $F_j=E_j-(E_1 \cup \cdots \cup E_{j-1})\subset E_j$ for all $j$, $\bigcup_{j=1}^{\infty} F_j \subset\bigcup_{j=1}^{\infty} E_j $ holds.

I have no idea to prove $\bigcup_{j=1}^{\infty} F_j \supset\bigcup_{j=1}^{\infty} E_j .$

答: Consider some $x\in \bigcup \limits_{j=1}^\infty E_j$. By definition $x\in E_j$ for some $j$ and we would like to say that $x\in F_j$. But what if eg. $x\in E_{j-1}$ as well? We need to make sure that the definition of $F_j$ doesn’t kill our element $x$. We can do this using the wellordering principle of the natural numbers: There is a minimal $j_0$ st $x\in E_{j_0}$ and $x\notin E_k$ for $k<j_0$. But then $x \in F_{j_0}$ and thus $x \in \bigcup_{j=0}^\infty F_j$.


问:
show that $(0.5)^x+(0.5)^{1/x}\le 1.$ if $x>0$
let $x>0$ show that
$$(0.5)^x+(0.5)^{1/x}\le 1.x>0,\tag{1}$$

Here is what I tried:let $f(x)=(0.5)^x+(0.5)^{1/x}$ so $f(x)=f(\frac{1}{x})$.si we only prove when $x\ge 1$.have $f(x)\le 1=f(1)$.it is sufficient to prove that
$$f'(x)=(0.5)^x\ln{0.5}-\dfrac{1}{x^2}\cdot(0.5)^{\frac{1}{x}}\ln{0.5}\le 0,x\ge 1$$
since $\ln{0.5}<0$,if we can show that
$$(0.5)^{\frac{1}{x}-x}\le x^2,x\ge 1$$
if we let $g(x)=x^2-(0.5)^{\frac{1}{x}-x}$,then $$g’(x)=2x+\ln{0.5}(0.5)^{\frac{1}{x}-x}\left(\dfrac{1}{x^2}+1\right)$$
it seem not easy to prove $g'(x)\ge 0,x\ge 1$ because $\ln{0.5}<0$.other words,this methods following more ugly!,so  How to prove this inequality $(1)$.Thanks.

答:
An alternative approach: Substitute $2^{-x} = u$. Then $0 < u < 1$ and $ 2^{-1/x} = 2^{\ln 2 /\ln u}$ so that
$$
\begin{align}
 &\, 2^{-x} +  2^{-1/x} \le 1 \\
\iff &\, u + 2^{\ln 2 /\ln u} \le 1 \\
\iff &\, \frac{(\ln 2)^2}{\ln u} \le \ln (1-u) \\
\iff &\, \ln(u) \ln (1-u) \le (\ln 2)^2 \, .
\end{align}
$$
Therefore we consider the function $g(u) = \ln(u) \ln (1-u) $ on the interval $(0, 1)$. $g$ is symmetric with respect to $u=1/2$, so that $g'(1/2) = 0$. The idea is now to show that $g$ is concave. Then it follows that $g$ has a maximum at $u=1/2$, i.e.
$$
\ln(u) \ln (1-u) = g(u) \le g(\frac 12) = (\ln 2)^2
$$
and we are done.

A straightforward calculation shows that the second derivative of $g$ is
$$
 g''(u) = -\frac{\log(1-u)}{u^2} -\frac{\log(u)}{(1-u)^2} - \frac{2}{u(1-u)} \, .
$$
Using the well-known inequality $\ln(y) \le y-1$ we have
$$
-\log(1-u) = \log \frac{1}{1-u} \le  \frac{u}{1-u}
$$
and
$$
-\log(u) = \log \frac{1}{u} \le  \frac{1-u}{u} \, .
$$
Substituting these estimates in $g''(u)$ gives $g''(u) \le 0$. So $g$ is concave on $(0, 1)$ and that concludes the proof.


问:
Show that the restriction of $f(x,y)=\frac{xy^2}{x^2+y^4}$ for every line passing through the origin is continuous.

Let $f(0)=0$ and $f(x,y)=\frac{xy^2}{x^2+y^4}$, when $(x,y) \ne0$. Show that $f$ is not continuous at the origin, but that the restriction of $f$ for every line passing through the origin is continuous.

It's easy to show that $f$ is not continuous at the origin by letting $y=\sqrt{x}$ for which $f(x,\sqrt{x}) = \frac12$ and letting $y=\sqrt{2x}$ for which $f(x,\sqrt{2x}) = \frac25$. However I'm not fully understanding what they mean by restricing $f$ for every line passing thourgh the origin? What does this actually mean?

答: It means that the restriction of $f$ to every set $\{(x,kx)\mid x\in\Bbb R\}$, as well as to the set $\{(0,x)\mid x\in\Bbb R\}$ is continuous.


问:
Finding a closed formula or an asymptotic bound
Can we find closed form of the formula

$$
\sum_{i=1}^{[\frac{n}{2}]} {n \choose i}(\sum_{j=i}^{n-i}{n-i\choose
j}-\frac{1}{2}{n-i \choose i})
$$

or, if not, is it possible to determine some asymptotic bound of it?

答: If I have understood the brackets correctly, then for $n \ge 1$ the values seem to be $0,1,6,25,90,301,\ldots$ which seems to be $$\tfrac12 3^n-2^n+\tfrac12$$ related to OEIS A000392


问:
Characteristic curves satisfy a straight line equation
let $\Omega$ be an open set in $\mathbb{R}^{n}$

let $F: \Omega \times \mathbb{R} \times \mathbb{R}^{n} \to \mathbb{R}$ be a smooth map defined as $ F(x,z,p) = \sum_{i=1}^{n} a_i(z) p_i$, where $a(z) = (a_1(z), ... , a_n(z)) \in \mathbb{R}^{n} $ is a smooth vector field.
Let $u$ be a smooth solution of $F(x, u(x), \nabla u(x)) =0$ for $ x \in \mathbb{R}^{n}$.

Let $x_0 \in \Omega$.
Let $(x_0, z_0, p_0) \in \Omega \times \mathbb{R} \times \mathbb{R}^{n}$.

Let $(X(t_0, x_0), z(t_0, x_0) ,p(t_0, x_0))$ be the associated bicharacteristic strip of $F$ passing  through the point $(x_0, z_0, p_0)$.
(The map $t \mapsto X(t, x_0)$ is called the characteristic curve)

Show that $ X(t, x_0)$ satisfies $ X(t, x_0) = x_0 + t a(u(x_0))$ in a straight line.

My attempt:
I am a beginner in PDE, having taken a course in ODE. I am stuck on this problem.
I have observed the following:
from hypothesis, $ \sum_{i=1}^{n} a_i(z) \frac{\partial u_i}{\partial x} = 0$.

Any inputs on how to proceed would be appreciated.

答:
I will show how to tackle the case $n=2$, see this Wikipedia article, "quasi-linear case" and "fully nonlinear case". So we have
$$
F(x, y, u, p, q) = a_1(u) p + a_2(u) q = 0
$$
with $p = u_x$ and $q = u_y$. To transform this PDE into differential equations, we introduce a parametrization of the solution of the form $u(x,y)$, $p(x,y)$, $q(x,y)$ with $x = x(t)$ and $y = y(t)$. Setting
$$
\dot x = F_p = a_1(u) , \qquad \dot y = F_q = a_2(u)
$$
gives us a system of ordinary differential equations. Now, let us compute the evolution of $u$ along these curves:
$$
\dot u = p\dot x + q\dot y = pF_p + qF_q = 0 \, ,
$$
according to the equation $F=0$.
Thus, $u = u(x_0, y_0)$ is constant along the integral curves $t\mapsto (x, y)$, as well as the derivatives $\dot x = a_1(u)$ and $\dot y = a_2(u)$ which are functions of $u$. In other words, the characteristic curves
$$
(x,y) = (x_0, y_0) + t\, a(u(x_0, y_0))
$$ are straight lines of the $x$-$y$ plane along which $u$ is constant. The generalisation for $n>2$ is straightforward.


问:
Let $f, g :(0,1) \to \mathbb{R}\,\,$ s.t. $\lVert g \rVert_2 =3$ and $\lVert f \rVert_1 = e\,\,$, show that $|\int_{0}^{1} g \sqrt{\log(f)}dx| \le 3$
Let $g\in L^2((0,1))$ s.t. $\lVert g \rVert_2 =3\,\,\,$ and $f \ge1, f \in L^1((0,1))$ s.t. $\lVert f \rVert_1 = e\,\,$, prove that:

$$ \Biggl|\int_{0}^{1} g \sqrt{\log(f)}\,dx \Biggr|\,\,\le\,\,3$$

My solution was:

$$ \Biggl|\int_{0}^{1} g \sqrt{\log(f)}\,dx \Biggr|\le \int_{0}^{1} \biggl|g \sqrt{\log(f)}\biggr|\,dx \le \biggl(\int_{0}^{1} |g|^2 \,dx\biggr)^{\frac{1}{2}}\biggl(\int_{0}^{1} \log(f)\,dx\biggr)^{\frac{1}{2}} =3\biggl(\int_{0}^{1} \log(f)\,dx\biggr)^{\frac{1}{2}}$$

the temptation is to say that $(\sup_{x\in(0,1)}|f|)\le\lVert f \rVert_1=e\,\,$ to conclude, but I don't know if it is true in this case. Any suggestion?

答:
It is not true that $|f(x)| \leq \|f\|_1$ for all $x$ so your argument does not hold.
Natural logarithm is concave so Jensen's inequaity gives $\int_0^{1} \log f \leq log \int_0^{1} f=1$.


问:
number of positive real root of $f(x)=x^4+2x^3-2 , x\in \Bbb{R}$,
$f(x)=x^4+2x^3-2 , x\in \Bbb{R}$,

(A) has two roots in $[0,\infty)$

(B) has three roots in $[0,\infty)$

(C) has no roots in $[0,\infty)$

(D) has a unique roots in $[0,\infty)$

How can I do this?I am stuck on it.

答: By Descartes rule of sign it has at most one positive real root.Now $f(0)=-2$  and $f(1)=1$.so it has a real root between $0$ and $1$.so (D) is the correct option.</p>


问:
How is max distributing over addition in this semi-formal proof?
There is a small proof (semi-formal I think) I am reading that as part of it has (the $\uparrow$ means the maximum function):

$t.p\uparrow t.r + (t.q - t.p)$

$=$  {distributive of addition over max, arithmetic} <=====?

$t.q\uparrow(t.r + (t.q - t.p))$

$\geq$  {$t.p \leq t.q$, monotonicity of max} (*)

$t.q\uparrow t.r$

Where I have the <=====?  it is the explanation given of how we move from each step to the next one, but I don't really understand it.

Could someone please explain how the distribution of adition over max worked in this case (or in general)?

Because what I can see is that:

$t.p\uparrow t.r + (t.q - t.p)$

there are 2 cases:

Case 1: $t.p > t.r$

So:

$t.p\uparrow t.r + (t.q - t.p)$

$\equiv$

$t.p + t.q - t.p$

$\equiv$

$t.q$

Case 2: $t.p < t.r$

So:

$t.p\uparrow t.r + (t.q - t.p)$

$\equiv$

$tr + t.q - t.p$

Since we can have either Case $1$ or Case $2$ the max is:

$t.p\uparrow t.r + (t.q - t.p) \equiv t.q\uparrow tr + t.q - t.p$

Which is what the proof states but it seems based on the wording that there is some fundamental operation that could be applied without needing to do this case analysis? I am looking for that explanation with this post

答:
For what's worth I figured out what that means:

$t.p\uparrow t.r + (t.q - t.p)$

$\equiv$

$t.p+ (t.q - t.p)\uparrow t.r + (t.q - t.p)$

$\equiv$

$t.p+ t.q - t.p\uparrow t.r + (t.q - t.p)$

$\equiv$

$t.q + t.p - t.p\uparrow t.r + (t.q - t.p)$

$\equiv$

$t.q \uparrow t.r + (t.q - t.p)$

So that should be the distributive property of addition over max


问:
Calculate E[T] w.r.t. Brownian motion
Let $(B_t)_{t\ge 0}$ be a standard Brownian Motion. Let
$$T:=\inf\left\{t\ge0:B_t=5t-2\right\}$$

Calculate $E[T]$.

My teacher said as a hint that I could use the fact that $E[T]$ is finite.
That would mean that according to Wald's Lemma, we know that 1. $E[B_T]=0$.
Also using Wald's 2nd Lemma it is known that $E[B_T^2]=E[T]$.

How can I also express $E[B_T]$ to solve this problem?

答: Since $T<\infty$ a.s.  and Brownian paths are continuous it follows that $B_T=5T-2$. Hence, $0=EB_T=E(5T-2)$ which gives $ET=\frac 2 5$.

Title: Calculate E[T] w.r.t. Brownian motion


问:
How can I construct a sequence of smooth functions $g_n$ s.t their integrals converge to zero?
Define $g_n(x)$ to be zero on $[0,1-\frac{1}{n}]$ and $g_n(x)=ax^2+bx+c$ on $[1-\frac{1}{n},1]$

I want the $g_n$ functions to be in $C^2[0,1]$ and statifies $g_n(0)=0$ and $g_n(1)=1$, $g_n'(0)=0$ and their integrals to converge to zero as suggested in the comments.

I solved the linear system in $a,b,c$ but to give: $a=n^2$, $b=-2n^2+2n$ and $c=(n-1)^2$

\begin{align}
\int_0^1 g_n(x)= \int_0^{1-\frac{1}{n}} 0 dx +  \int_{1-\frac{1}{n}}^1 n^2x^2 -2n(n+1)x+ (n-1)^2 dx = \dfrac{n^2-3n+3}{3}-\dfrac{n^3-3n^2+3n-1}{3n}=\dfrac{3}{n}
\end{align}

so Let $a_n=\frac{3}{n}$

\begin{align}
\lim_{n \to \infty} a_n=\dfrac{n^2-3n+3}{3}-\dfrac{n^3-3n^2+3n-1}{3n} =\lim_{n \to \infty} \frac{3}{n} =0
\end{align}

EDIT: Updated the body of the question

答: As it was pointed out in the comments, the functions $g_n$ are not even continuous en $[0,1]$. Regarding the second part, it is true that $\int_0^1 g_n(x) dx \to 0$ and your justification is correct.


问:
Find $\mathcal{O}_{\mathscr{L}(a)}$ so that $a_n\sim\sqrt{1+\frac{1}{1+a_n}}+\mathcal{O}_{\mathscr{L}(a)}(\frac{1}{n})\,{\rm as}\,n\rightarrow\infty$

Given a recursion $a_{n+ 1}= \sqrt{1+ \dfrac{1}{1+ a_{n}}}$ with $a_{1}> 1.$ We have that
$$a_{n}\sim\sqrt{1+ \frac{1}{1+ a_{n}}}+ \mathcal{O}\left ( \frac{1}{n} \right )\,{\rm as}\,n\rightarrow\infty$$
Source: It's an my own problem that I posted before and that received many solutions.

Call $\lim a_{n}= a,$ if I want to give an asymptotic analysis for $a_{n+ 1}- a_{n},$ that's no good since $a\not\in\mathbb{Q}.$ I think about Groebner Basis, is there way to find $\mathcal{O}_{\mathscr{L}\left ( a \right )}\left ( \dfrac{1}{n} \right )$ with $a$ is a root of the irreducible polynomial $\mathscr{L}\left ( a \right ),$ how can we do that ?

Body:
Note that
$$
\left| {a_{n + 1}  - a_n } \right| = \frac{{\left| {a_n  - a_{n - 1} } \right|}}{{(1 + a_n )(1 + a_{n - 1} )\left( {\sqrt {1 + \frac{1}{{1 + a_n }}}  + \sqrt {1 + \frac{1}{{1 + a_{n - 1} }}} } \right)}} \le \frac{{\left| {a_n  - a_{n - 1} } \right|}}{8}
$$
since $a_n>1$ for all $n\geq 1$. Thus,
$$
\left| {a_{n + 1}  - a_n } \right| \le \frac{{\left| {a_2  - a_1 } \right|}}{{8^{n - 1} }} \le \frac{{4\sqrt 6  + 8a_1 }}{{8^n }},
$$
i.e.,
$$
a_{n + 1}  = a_n  +\mathcal{O}\!\left( {\frac{1}{{8^n }}} \right).
$$


问:
How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$?
This is a multiple choice question, where using calculators is not allowed. Candidates have, on average, two minutes to solve the problem.

PROBLEM:

How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$?

CHOICES:

(A) $4$ (B) $5$ (C) $6$ (D) $7$ (E) $8$

What I have tried was not feasible:

Using differentiation, I found that the curve is bounded between $y=\frac{9}{8}$ and $y=-2$.

When $x<-8$, $\frac{1}{4}x<-2$ , and when $x>\frac{9}{2}$, $\frac{1}{4}x>\frac{9}{8}$. So, any solution to the equation $y=\sin(x)+\cos(2x)=\frac{1}{4}x$ must lie in the interval $(-8,\frac{9}{2})$.

I do not know how to proceed from these two points. Even I know, that takes a long time.

Also, I do not know if using the fact that the curve is periodic will really help.

By Desmos, I took a look to the graph, the line cuts the curve $7$ times. So the correct choice is $D$.

Any help on how to solve this would be appreciated. THANKS!

答:
The function is periodic with period $2\pi$ and achieves local extrema

$$y' = \cos x - 2\sin 2x = 0 \implies \cos x (1-4\sin x) = 0$$

at

$$\pm\sqrt{1-\cos^2x}+2\cos^2x-1\Bigr|_{\cos x = 0} = 0,-2$$

and

$$\sin x + 1-2\sin^2x\Bigr|_{\sin x = \frac{1}{4}} = \frac{9}{8}$$

which means the only interval we can expect intersections is on $\left[-8,\frac{9}{2}\right]$. The function has four extrema per period and the interval contains

$$\frac{25}{4\pi} = \frac{8\pi+(25-8\pi)}{4\pi} = 2 + \epsilon$$

two whole periods by estimating since $\epsilon < \frac{1}{12}$. $2\cdot 4 = 8$ turn arounds means $7$ intersections between them.


答:
Denote $T : L^p [0,1] \to L^p [0,1]$ the integral operator
$$
(Tf)(u) = \int_0^u f(t)\, dt.
$$
Consider an operator $T^n$. Its norm can be evaluated as follows:
$$
\begin{align*}
||T^n f||_p &= \sqrt[p]{\int_0^1 dx \, \left|\int_0^x dt_n\dots\int_0^{t_2}f(t_1)\, dt_1\right|^p} \leq \sqrt[p]{\int_0^1 dx \, \left[\int_0^x dt_n\dots\int_0^{1}|f(t_1)|\, dt_1\right]^p}\\
&\leq \sqrt[p]{\int_0^1 dx \, \left[\int_0^x dt_n\dots\int_0^{t_3}||f||_p \, dt_2\right]^p} = ||f||_p \sqrt[p]{\int_0^1 dx \, \left[\int_0^x dt_n\dots\int_0^{t_3}\, dt_2\right]^p}\\
&= ||f||_p \sqrt[p]{\int_0^1 dx \, \left[\frac{x^{n-1}}{(n-1)!}\right]^p} \leq \frac{||f||_p}{(n-1)!} \sqrt[p]{\int_0^1 dx \, x^{n-1}} \leq \frac{||f||_p}{(n-1)!},
\end{align*}
$$
so $||T^n|| \leq \frac{1}{(n-1)!}$. By the way, I used Holder's inequality in the second inequality. By Gelfand's formula for spectral radius $r(T)$
$$
r(T) = \lim_{n \to \infty} \sqrt[n]{||T^n||} \leq \lim_{n \to \infty} \frac{1}{\sqrt[n]{(n-1)!}} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}} = 0.
$$
In particular, each eigenvalue of $T$ (if they exist in the first place) is zero.

Suppose we have
$$
\int_0^u f(t) \, dt = 0.
$$
for some $f \in L^p [0,1]$. Then for any interval $X \subset [0,1]$ we have $\int_X f(t) \, dt = 0$. Thus $\int_X f(t) \, dt = 0$ for any measurable $X \subset [0,1]$, i.e. $f = 0$ in $L^p [0,1]$.

Title: Eigenvalues of integral operator

Body:
Let $f \in L^{p}[0,1]$ for $p \in (1,\infty)$ and consider the operator $Tf(u) = \int_{0}^{u} f(t) dt$. We want to show that $T$ has no eigenvalues. An eigenvalue $\lambda \in \mathbb{R}$ means that there is a non-trivial solution to the equation
$$
\lambda f(u) = \int_{0}^{u} f(t) dt
$$
for all $t \in [0,1]$.

Note that this is easy to do when $f$ is assumed to be continuous. We differentiate both sides to find the $f$ has to be something exponential and realize that $f(0) = 0$ (thus no solution). However, when $f$ is simply in $L^{p}$ but not continuous we cannot do this. My thought is to use the density of continuous functions in $L^{p}$ to do this, but I'm not able to formalize the argument.

My "argument" flows like this: If for fixed $\lambda \in \mathbb{R}$ has the property that
$$
\int_{0}^{u} f(t) dt - \lambda f(u) = 0
$$
for all $u \in [0,1]$, this implies that there is a continuous $g$ such that $\int_{0}^{u} g - \lambda g = 0$, which is a contradiction. However, I'm not sure how to prove the existence of such a $g$. Is it even true?


问:
Calculate expectation and variance of smooth function applied to brownian motion
I applied a smoothing function to a Brownian equation and obtained a stochastic differential equation by using Ito's lemma. The smoothing function is $e^{Bt}$.

How do I get the expected value and variance of this function? Just looking for the required approach rather than a full fledged solution.

答:
It is my question also. I found a look like question  in "An Algorithmic Introduction to
Numerical Simulation of
Stochastic Differential
Equations" by prof. Desmond J. Higham

https://doi.org/10.1137/S0036144500378302

and it is about the expected value of $e^{t+0.5B_t}$ the research says that $\mathbb{E(x)}=e^{\frac 98 t}$I tried to find expect vale by direct definition but got nothing. I tell you about my experience. I use Ito difference formula to understand $\mathbb{E(x)}$.
take $g(t,x)=e^{t+0.5x}$ and $$dg=\frac{\partial g}{\partial t}dt+\frac{\partial g}{\partial x}dx+\frac12\frac{\partial^2 g}{\partial x^2}(dx)^2\\dg=
e^{t+0.5x}dt+0.5e^{t+0.5x}dB_t+\frac12(0.5)(0.5)e^{t+0.5x}(dB_t)^2=\\\frac98e^{t+0.5x}dt+0.5e^{t+0.5B_t}db_t\to \mathbb{E}(g)=e^{\frac98t}$$  so (I think similarly) when we want to find expected value for $e^{B_t}$ we can take $$g(t,x)=e^{x}\\It\hat{o} \space diff \space dg=0dt+1e^{B_t}dB_t+\frac{1}{2}e^{B_t}dt\\$$so it shows maybe ($\mathbb{E}(x)=e^{\frac12t})$ Iam not sure that is right approach or not, but hope it can help you.


问:
Use epsilon-delta definition of limits to show that $\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$ and $\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$
Use epsilon-delta definition of limits to show that

$$\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$$
$$\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$$

We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{\left(x-1\right)^{2}+\left(y-2\right)^{2}}<\delta \implies \left|\left(3x+2y-1\right)-6\right|<\epsilon)))$$

Taking $\left|x-1\right|,\left|y-2\right|<\delta$ follows :
$$\left|\left(3x+2y-1\right)-6\right|\le 3\left|x-2\right|+\left|2y-1\right|<3(\delta+1)+(2\delta+3)$$
$$\left|\left(3x+2y-1\right)-6\right|<5\delta +6$$

Taking $ \delta \le{(\epsilon -6)}/5$ shows the claim.

We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{x^2+y^2}<\delta \implies \left|x^2+y^2 \right|<\epsilon)))$$
$$\left|x^2+y^2 \right|=x^2+y^2$$

Taking $ \delta \le\sqrt{\epsilon }$ shows the claim.

I want to know if my solution is true,so if you have an alternative solution that would be nice to see that,but please first check mine.

答:
The latter one is correct, but the former is not. A big hint is that the choice $\delta \le (\varepsilon - 6) / 5$, which is negative(!) when $\varepsilon < 6$. Your choice of $\delta$ should always be positive, so that at least one $(x, y)$ satisfies $0 < \|(x, y) - (x_0, y_0)\| < \delta$.

Instead, consider that
$$|(3x + 2y - 1) - 6| = |3(x - 1) + 2(y - 2)| \le 3|x - 1| + 2|y-2|.$$
So, if we take $\delta = \varepsilon / 5$, then
\begin{align*}
0 < \sqrt{(x - 1)^2 + (y - 2)^2} < \delta &\implies |x - 1|, |y - 2| < \frac{\varepsilon}{5} \\
&\implies |(3x + 2y - 1) - 6| < 3\frac{\varepsilon}{5} + 2\frac{\varepsilon}{5} = \varepsilon.
\end{align*}


问:
How to find $δ$ given $\varepsilon = 1$ of $\lim_{x \to 1} f(x)$, $f(x)= 1/(x(x-2))$?
I have been asked to find a $\delta$ given $\varepsilon=1$ for $f(x) = \frac{1}{x(x-2)}$, and I'm stuck right after setting up the $\vert f(x)-L\vert$ portion of the definition. I ended up with an inequality that looks like this:

$$-1<\frac{1}{x(x-2)}+1<1$$

Then I subtracted 1 from both sides and am left with with this

$$-2<\frac{1}{x(x-2)}<0$$

This is where I'm stuck.. I've been told that I can do a reciprocal if both sides of the inequality are positive, but since one side is negative I'm not certain if it would be the correct step. Apologies for formatting. first time poster.

答:
Regarding your last inequality:
$$-2 < \frac{1}{x(x-2)} < 0$$
Note that this is equivalent to
$$\begin{cases}\begin{align}
\frac{1}{x(x-2)} &< 0 \\
\frac{1}{x(x-2)} &> -2
\end{align}\end{cases} \implies
\begin{cases}
x \in (0,2) \\[1mm]
x \in \left(-\infty \:,\:0\right)\cup \left(\frac{2-\sqrt{2}}{2},\:\frac{2+\sqrt{2}}{2}\right)\cup \left(2,\:\infty \:\right)\end{cases}$$
Now, find the intersecti0n of the two results:
$$x\in \left(\frac{2-\sqrt{2}}{2},\:\frac{2+\sqrt{2}}{2}\right)$$


问:
How to prove $G_2$ is a normal subgroup of $G_1\times G_2$?
I am studying abstract algebra and in one of my previous midterms there was a question that I can not solve. Now I have final exam tomorrow and want to learn the answer to that question. Here it is:

Can anyone help me with the solution of this question? My answer is there, but it would be very nice if you assume there is no answer and explain the answer to the question for part (i) and (ii), I really want to understand it. Thank you in advance

答:
Hints. (i) $(g_1,g_2)(e,g)(g_1,g_2)^{-1}=(g_1,g_2)(e,g)(g_1^{-1},g_2^{-1})=(g_1eg_1^{-1},g_2gg_2^{-1})$ $=$ $(e,g_2gg_2^{-1})$ $\in$ $\{e\}\times G_2$ for any $g_1\in G_1$ and $g,g_2\in G_2$.

(ii) Define $\varphi:G_1\times G_2\to G_1$ by $\varphi(g_1,g_2)=g_1$. Note that $\varphi$ is a surjective group homomorphism, $\ker\varphi=\{e\}\times G_2$ and apply the fundamental theorem of isomorphism for groups.


问:
If $(a^{n}+n ) \mid (b^{n}+n)$ for all $n$, then $ a=b$
I happened to receive this from my friend.

Let $a,b \in \mathbb{N}$, such that $a^{n}+n \: \bigl|\: b^{n}+n$ for all $n \in \mathbb{N}$. Prove that $a=b$. How do we proceed?

答:
Claim if our hypothesis holds, $a \equiv b \ (\text{mod}\ p)$ for any prime $p$.

Proof:

Find $n$ so that $n \equiv -a \ (\text{mod}\ p)$ and $n \equiv 1 (\text{mod}\ p-1)$ ( we can do this by Chinese Remainder Theorem).  Then

$$a^{n} + n \equiv a^1 + n = a - a = 0 (\text{mod} \ p)$$

Therefore since $a^n + n \mid b^n + n$, $b^n + n \equiv 0 (\text{mod}\ p)$

But

$$b^n + n \equiv b^1 + n \equiv b - a (\text{mod}\ p)$$

therefore $b \equiv a \pmod p$.

Our result now follows by picking any $p > b$.

NOTE by BD $\;$  This solution has been posted before in at least a few well-known math forums,  e.g. see my sci.math post on April 4,2006, and see Rust's post on AoPS, July 19, 2009. It also appeared in at least one other forum much more recently (alas, I can't recall which one). Almost surely, by now the problem and solution is listed in various problem collections, so it should be considered somewhat well-known.

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: Dirichlet’s Theorem: Every proper arithmetic sequence contains a prime.</p>

LastEditorDisplayName: anon


问:
Given a continuous monotone function $f$, is $f^{-1}(x)$ a connected set?
Given a continuous monotone function $f: \mathbb{R} \rightarrow \mathbb{R}$, is it true that for any point $x \in \mathbb{R}$, $f^{-1}(x)$ must also be connected?

The monotonicity is defined as a non strict monotonicity, and therefore $f^{-1}(x)$ is not necessarily a point.

A simmilar property has already been proven false for $\mathbb{R}^m$ in general in the question below, but I want to know if it is true for the case of $\mathbb{R}$, and in the case of a point $x$ insted of a subset $C \subset \mathbb{R}$.

Continuous monotone maps and pre-image of connected sets.

答:
Yes, it is true. Suppose otherwise. Since, in $\Bbb R$, being connected is the same thing as being an interval, if the statement was false, there would be $a,b\in f^{-1}(x)$, with $a<b$, and there would be $c\in(a,b)$ with $c\notin f^{-1}(x)$. But then $f(c)>x$ or $f(c)<x$. Since $f(a)=f(b)=x$, this contradicts the assumption that $f$ is monotonous.

Note that, as you were told in the comments, continuity is not needed to prove this.


问:
Сardinality of the set of increasing functions from Z to Z?
Could you show me how to "calculate" the cardinality of the set of increasing (not necessarily strictly) functions  $\ f: \mathbb{Z} \rightarrow \mathbb{Z}$ ?</p>

答: Let $g\left(n\right)=f\left(n \right)-f\left(n-1\right)$. Then $g:\mathbb{Z}\mapsto\mathbb{Z}_+$ that. by varying $f$, can be any arbitrary sequence of non-negative integers. Knowing $g$ and $f\left(0\right)$ determines $f$, uniquely. So you can look at the cardinality of the set of possible pairs $\left(g,f\left(0\right)\right)$. The cardinality should be the cardinality of $\mathbb{R}$.</p>


问:
Conditional probability lie detector
A lie detector returns a positive result when a person is lying in
90% of all cases. Unfortunately, it also results a positive result when a person
is telling the truth in 20% of all cases. Statistically, it is estimated that 10% of
all test subjects lie.
A randomly chosen test subject takes a test and obtains a negative result.
What is the probability that the subject told the truth?

So this was a 2 part question in which I have taken $A=$ Positive Result and $B=$ Subject is Lying:
I've then calculated $P(B)=0.1,$ $P(A|B^c)=0.2,$ $P(A|B)=0.9,$ $P(B^c)= 0.9,$
and  $P(B|A)=1/3.$
I was thinking of using Bayes' formula and using the law of total probability to figure out $P(B^c|A^c)$ but I am struggling with this. If anyone can point me in the right direction it would be appreciated.

答:
Hint
\begin{equation}
\begin{array}{c|c|c}&\text{positive}&\text{negative}\\
\hline \text{non liars}&0.18&0.72\\
\hline \text{liars}&0.09&0.01
\end{array}
\end{equation}

Title: Conditional probability lie detector


问:
Exponential of upper triangular matrix
Consider the matrix
$$
A=\left(\begin{matrix}a&b\\0&a\end{matrix}\right),\qquad
A^n=\left(\begin{matrix}a^n&b(na^{n-1})\\0&a^n\end{matrix}\right)
$$
The exponential series is
$$
\exp(X)=\sum_{k=0}^\infty \frac{1}{k!}X^k
$$

Thus take the formal exponential of $A$ I get
$$
\exp(A)=\sum_{k=0}^\infty \frac{1}{k!}\left(\begin{matrix}a^k&b(ka^{k-1})\\0&a^k\end{matrix}\right)
$$
Since $ka^{k-1}$ is just the derivative of $a^k$, and $\frac{d}{dx}e^x=e^x$, I get
$$\tag{*}
\exp(A)=\left(\begin{matrix}e^a&be^a\\0&e^a\end{matrix}\right)
=e^a\left(\begin{matrix}1&b\\0&1\end{matrix}\right)
$$

An easier computation that gives some consistency would be
$$
\begin{align}
\exp\left(\begin{matrix}a&b\\0&a\end{matrix}\right)
&=\exp\left[\left(\begin{matrix}a&0\\0&a\end{matrix}\right)
+\left(\begin{matrix}0&b\\0&0\end{matrix}\right)\right]\\
&=\exp\left(\begin{matrix}a&0\\0&a\end{matrix}\right)\exp \left(\begin{matrix}0&b\\0&0\end{matrix}\right)
=\left(\begin{matrix}e^a&0\\0&e^a\end{matrix}\right)
\left(\begin{matrix}1&b\\0&1\end{matrix}\right)
\end{align}
$$
since in exponential expansion of the second factor only the first two terms are relevant.

(Q1) Makes this computation any sense?
Is it generalisable for the Jordan normal form?

(Q2) If we look at
$$
A=\left(\begin{matrix}a&b\\0&a\end{matrix}\right)
=\left(\begin{matrix}a&0\\0&a\end{matrix}\right)\left(\begin{matrix}1&b/a\\0&1\end{matrix}\right)
$$
makes the following any sense, or can we extract anything interesting from it?
$$
e^a\left(\begin{matrix}1&b\\0&1\end{matrix}\right)
=\exp A = \left(\begin{matrix}e^a&0\\0&e^a\end{matrix}\right)^{\left(\begin{matrix}1&b/a\\0&1\end{matrix}\right)}
$$

答:
(Q1), yes this makes sense, since $\left(\begin{matrix}a&0\\0&a\end{matrix}\right)$ and $\left(\begin{matrix}0&b\\0&0\end{matrix}\right)$ commute.

(Q2),  $\left(\begin{matrix}e^a&0\\0&e^a\end{matrix}\right)^{\left(\begin{matrix}1&b/a\\0&1\end{matrix}\right)}$ makes no sense.

Title: Exponential of upper triangular matrix


问:
Interpretation of a question on probabilty
I'm not sure this type of question is for this SE but I really need an answer possibly just as a comment.

''For many years, Basil and Rosemary have spent long weekends (4 days each) at the resort of Wetsun. They have observed that two sunny days in three are followed by another sunny day, while half of all the wet days are followed by another wet day. Days are either wet or sunny.''

I'm not sure if the question means:

"two sunny days in three [days]" i.e. of three days, if two are sunny, then the next day must be sunny, for example:

SSW->S

SWS->S

WSS->S

where -> means followed by.

"two sunny days in three [sunny days]" i.e. two thirds of sunny days are followed by as sunny day

答:
Here is the original text:

They have observed that two sunny days in three are followed by another sunny day, while half of all the wet days are followed by another wet day. Days are either wet or sunny.

Here is my revision, staying as close to the original structure as possible:

They have observed that, on average, every two out of three sunny days are followed by another sunny day, while on average, half of the time, a wet day is followed by another wet day.  Days are either wet or sunny.

I would state it like this:

If on a given day it is sunny, the probability that the next day will also be sunny is $2/3$.

If on a given day it is wet, the probability that the next day will also be wet is $1/2$.

The language is strangely phrased, but that is very clearly the intent.


问:
Can the tangent line be defined independently of the derivative?
The graph of the function $f:x \mapsto x^{1/3}$ has a 'vertical tangent' at $x=0$:

Although this idea is certainly geometrically sound, from what I understand the tangent line is defined by the derivative, not vice versa. In other words, the tangent line to a function at the point $(a,f(a))$ is simply the line given by the equation
$$
y - f(a) = f'(a)(x-a) \, ,
$$
where $f'(a)$ is of course defined as a limit. Since $f'(0)$ does not exist in this case, I'm unsure if we can truly say that the graph has a vertical tangent. The intuitive idea of a tangent 'just touching' the curve breaks down when we consider, for instance, the graph of a linear function, where the tangent touches the graph of the function itself at infinitely many points. Nevertheless, I have heard people say that the tangent line is fundamentally a geometric concept. Although the slope of the tangent line 'agrees' with the derivative if the derivative exists, there are instances where the tangent line is a meaningful concept even when the derivative doesn't exist. If this be the case, then what is the formal definition of a tangent?

答: You won't find a definition of the tangent line that is completely independent of the general concept of derivatives, since they are so intimately connected.  But from the point of view of differential geometry, you can most certainly have vertical tangent lines. To this end, we have to slightly shift our point of view: we're not looking for tangent lines to the graph of a function, but to some smooth set of points, which happens to be easily described as the graph of a function. More rigorously, we're interpreting the graph as a smooth submanifold of $\mathbb R^2$. In this case, the tangent line at $p$ can be parameterized as $t_p(\tau)=p+\tau v$, where $v$ is a nonzero vector tangent to the manifold at $p$, in this case $(0,1)$, for instance. Of course, to find such a vector, we'll end up using derivatives anyway: we'd parametrize the manifold, say by the smooth curve $\gamma:\tau\mapsto(\tau^3,\tau)$, and then take its derivative at $\tau=\gamma^{-1}(p)$ (since $\tau=\gamma^{-1}(p)$ yields $p$ when inserted into $\gamma$). This derivative is a vector tangent to the parameterized curve/manifold at $p=0$.


答:
Hint:

How many functions are there from $\mathbb{Z}$ to $\mathbb{Z}$? We note that there are $|\mathbb{Z}|$-many values such a function can take at each point $z$ in $\mathbb{Z}$.

How many monotonic functions are there from $\mathbb{Z}$ to $\mathbb{Z}$? We note that  there are still $|\mathbb{Z}|$-many values such a function can take at each point $z$ in $\mathbb{Z}$ -- since there are still $|\mathbb{Z}|$-many possible values in the interval $[f(z -1 ), \infty)$ where the value of $f(z)$ must fall.

Title: Сardinality of the set of increasing functions from Z to Z?

Body: Could you show me how to "calculate" the cardinality of the set of increasing (not necessarily strictly) functions  $\ f: \mathbb{Z} \rightarrow \mathbb{Z}$ ?</p>


问:
Homework hint: use dot product to show inscribed angle of a semicircle is a right angle
I'm currently stuck on a homework problem from MIT OCW 18.02 multivariate calculus.

The problem is in a section on dot products, and asks to use vector methods (without components) to show that the inscribed angle of a semicircle is a right angle.

I have been toying around with it the past couple of days. The first thing I did was sketch it out, and I quickly came to a solution by summing up the angles in my sketch. (I drew two triangles and determined the relationship between the angles, which led me to deduce the inscribed angle was $\frac{\pi}{2}$)

But that solution doesn't feel in keeping with the rest of the problems, which makes me wonder what I missed.

I know I could just peek the answer, but I also know that wouldn't really help. I would prefer a gentle nudge in the right direction instead. Can you provide a hint that may point me in the direction I need to think?

答:
Hint :

Since vectors encode same information as Geometry, we make use of same triangles in (your) Euclidean geometry proof.

Fix the center of semicircle as the origin. Let $A,B$ be the position vectors of end points of the base diameter and let $P$ be the position vector of any point on the semicircle. We have to prove
$$(P-A)\cdot (P-B)=0$$

Suppose $A=\vec{a}$, what would be $B$? Let $P=\vec{r}$.
Now that we know two sides of each triangle, we can find out the third.

Also note that $|OP|=|OA|=|OB|$. Now we can find out $PA \cdot PB$ and see if it evaluates to zero.


问:
Higher commutators in rings
In Herstein's "Noncommutative rings" it is proved that any rng $R$ such that for every $x,y\in R$ there exists an integer $n(x,y)>1$ such that $$(xy-yx)^{n(x,y)}=(xy-yx)$$ must be commutative. At the end of the section he says that the previous proof can be generalized replacing $[x,y]=xy-yx$ by higher commutators, e.g. $[[x,y],z]$, $[[[x,y],z],w]]$ and so on. I think I managed to modify his proof in order to prove this fact, but then he says that if $R$ does not contain nil ideals, then $R$ is even commutative. I'm struggling proving this latter fact. Let's take into account the commutator $[[x,y],z]$, then my problem is the following one:

If $R$ is a rng satisfying $[[x,y],z]=0$ for all $x,y,z\in R$ and $R$ does not contain nil ideals, then $R$ is commutative.

I was trying to prove that the ideal generated by all commutators $[x,y]$ with $x,y\in R$ is nil, but that does not seem to be the right path. Do you have any suggestion?

答:
We can prove this when $R$ is free of $2$-torsion:

$[[x,y],z]=0$ means that any commutator $[x,y]$ is in the center $Z(R)$.

$[x,[y,zw]]=[x,[y,z]]w+z[x,[y,w]]+[x,z][y,w]+[y,z][x,w]$ implies that
$$[x,z][y,w]+[y,z][x,w]=0,$$
in particular, when $R$ is free of $2$-torsion,
$$[x,z][x,w]=0.$$

Now consider the ideal $I_a$ of $R$ generated by the elements $[a,x]$ for fixed $a\in R$ and all $x\in R$. Elements of $I_a$ are sums of the form $y[a,x]z$ (where we can have "$y=1,z=1$"), and so by items 1. and 2. we get $I_a^2=0$; in particular $I_a$ is a nil ideal, so $I_a=0$, hence $[a,R]=0$ and $a\in Z(R)$. Since this is true for all $a\in R$ we get $R=Z(R)$.

Note that the result is true also more in general for semiprime rings, i.e., rings without nonzero nilpotent ideals.

Title: Higher commutators in rings


问:
Calculate the residue of $f(z) = \frac {e^{-z}} {(z+2)^3} $
I'm currently studying integral solving methods using the Residue Theorem. However, I sometimes have problems finding a way to calculate the residue(s). For example, what do I do when a function includes the exponential function (or any other function whose Taylor series is commonly known)?

After a while, I found a solution that I added below this question. In general, I thought that it might be helpful to others when they are dealing with solving integrals using the residue theorem for functions that involve the exponential function (or any variants of it).

My initial problem was:

How do I calculate the residue of $f(z) = \frac {e^{-z}} {(z+2)^3} $ at $z = -2 ?$

With the solution to this problem, I was then able to calculate the integral using the residue theorem:
$$
\int_{|z|=3} f(z) dz =  ?
$$

答:
That's a fair solution. Other than that you could also calculate the residue as follows:
We have
\begin{align}
\operatorname{Res}_{-2}f = \frac{1}{(3-1)!}\frac{\mathrm{d}^2}{\mathrm{d}z^2}\left((z+2)^3\cdot f(z)\right)\bigg|_{z = -2},
\end{align}
since we have a pole of order $3$ at $z = -2$. Due to
\begin{align}
\frac{1}{(3-1)!}\cdot\frac{\mathrm{d}^2}{\mathrm{d}z^2}\left((z+2)^3\cdot f(z)\right) = \frac{1}{2!}\cdot\frac{\mathrm{d}^2}{\mathrm{d}z^2} e^{-z} = \frac{e^{-z}}{2},
\end{align}
we find
\begin{align}
\boxed{\operatorname{Res}_{-2}f = \frac{e^{2}}{2}.}
\end{align}

Edit:

Here I used the general formula

\begin{align}
c_{-1} = \operatorname{Res}_af = \frac{1}{(k-1)!}\cdot \frac{\mathrm{d}^{k-1}}{\mathrm{d}z^{k-1}}\left((z-a)^k\cdot f(z) \right) \bigg|_{z=a},
\end{align}
if $a$ is a pole of order $k$ of $f$.
What this formula does is basically extracting the coefficient $c_{-1}$ of the Laurent series which by definition is your residue.


问:
A evolutionary triple relating to Sobolev space
In Brezis's functional analysis(p.291), it says the following

$$W_0^{1,p}(\Omega) \subset L^2(\Omega) \subset W^{-1, p'}(\Omega) \quad \mbox{if} \quad 2N/(N+2) \le p \le 2$$

whether $\Omega$ is bounded or not bounded.  As far as I know, most of Sobolev's inequalities or embeddings apply to bounded domains. Would you explain why the first inclusion holds? What Sobolev inequalities should we use to prove that?

答:
The space $W^{1,p}_0$ is continuously embedded into $L^q$ for $\frac1q=\frac1p-\frac1N$ (Thm 9.9 in Brezis).
The value of $q$ is larger than $2$ iff $\frac12\ge\frac1q=\frac1p-\frac1N$ or equivalently $p\ge \frac{2N}{N+2}$. Then we know: $W^{1,p}\subset L^q \cap L^p$,
and since $p \le 2\le q$, $W^{1,p}\subset L^2$ by Hoelder inequality.


问:
Does the category of semisimplicial Kan complexes form a category of fibrant objects?
It is known that the category of Kan complexes form a category of fibrant objects, as remarked here. We obtain obvious notions for fibrations and weak equivalences in the category of semisimplicial sets, being the Kan fibrations and the maps inducing homotopy equivalences. Does the full subcategory of semisimplicial sets, consisting of the semisimplicial Kan complexes, give a category of fibrant objects?

Or, maybe more general, does the forgetful functor from simplicial sets to semisimplicial sets induce a model structure on semisimplicial sets? I've read from various resources that this is not necessarily the case, but to which extend is this structure preserved by the forgetful functor? That is, which of the axioms of a model category are satisfied, and which aren't, by considering this structure inherited from simplicial sets? What is the standard literature on this?

There are some discussions like this, or this, but these do not entirely answer the question with great certainty.

答:
For clarity, let's fix precise definitions.
A Kan fibration of semisimplicial sets is a morphism that has the right lifting property w.r.t. the horn inclusions $\Lambda^n_k \hookrightarrow \Delta^n$, where $\Delta^n$ here means the semisimplicial set represented by $[n]$.
A weak homotopy equivalence of semisimplicial sets is a morphism whose geometric realisation is a homotopy equivalence.

In brief:

There does not exist a model structure on the category of semisimplicial sets where the weak equivalences are the weak homotopy equivalences.

There does not exist a full subcategory of semisimplicial sets that contains $\Delta^0$ as well as non-discrete semisimplicial sets and admits the structure of a category of fibrant objects where the weak equivalences are the weak homotopy equivalences and pullbacks of fibrations are preserved by the inclusion functor.
(This is not quite an answer to your question since $\Delta^0$ is not a Kan-fibrant semisimplicial set.)

Let $i^*$ be the forgetful functor from simplicial sets to semisimplicial sets.
It has a left adjoint $i_!$ and a right adjoint $i_*$.
It is clear that $i_! (\Lambda^n_k \hookrightarrow \Delta^n)$ is (isomorphic to) the corresponding horn inclusion in simplicial sets, so a morphism $X \to Y$ of simplicial sets is a Kan fibration if and only if $i^* (X \to Y)$ is a Kan fibration of semisimplicial sets.
It is also clear that a morphism $X \to Y$ of semisimplicial sets is a weak homotopy equivlence if and only if $i_! (X \to Y)$ is a weak homotopy equivalence of simplicial sets.
This sounds good, but notice that the functors involved in the two observations are going in opposite directions!
That is basically why we can't use the usual machinery to transfer the model structure from simplicial sets to semisimplicial sets.

In fact, as I claimed above, there is simply no way of constructing a model structure or a category of fibrant objects with the desired properties.

The comments here explain why there is no possible model structure on the category of semisimplicial sets where the weak equivalences are the weak homotopy equivalences defined above.
I reproduce the argument.
Suppose there is such a model structure.
Let $p : X \to Y$ be a trivial fibration in the putative model structure.
As in any model category, the class of trivial fibrations is closed under pullback, so for each $n$-simplex $y : \Delta^n \to Y$ we obtain a trivial fibration $p_y : X_y \to \Delta^n$.
But $\Delta^n$ has no $m$-simplices for $m > n$, so $X_y$ also has no $m$-simplices.
In the case $n = 0$, $p_y : X_y \to \Delta^0$ must therefore be an isomorphism.
By induction on $n$, we deduce that $p : X \to Y$ is an isomorphism on the $n$-skeleton for every $n \ge 0$, hence $p : X \to Y$ is an isomorphism of semisimplicial sets.
Thus every trivial fibration in the putative model structure is an isomorphism.
This implies every morphism is a cofibration, which would mean every pushout is automatically a homotopy pushout – but e.g. the pushout of $\Delta^0 \leftarrow \Delta^0 \amalg \Delta^0 \rightarrow \Delta^0$ is not a homotopy pushout.
Therefore there is no such model structure.

The above argument does not, prima facie, preclude the possibility that there is some full subcategory of semisimplicial sets that admits the structure of a category of fibrant objects containing $\Delta^0$ and where the weak equivalences are the weak homotopy equivalences and the inclusion preserves pullbacks of fibrations.
Since $\Delta^0$ is assumed fibrant, we can apply the earlier argument to deduce that every trivial fibration (of fibrant semisimplicial sets) is an isomorphism of 0-skeleta.
This is already problematic.
Let $X$ be a fibrant semisimplicial set.
It has a path object, i.e. $(P, i, p_0, p_1)$ where $P$ is a fibrant semisimplicial set, $i : X \to P$ is a weak homotopy equivalence, $p_0, p_1 : P \to X$ are morphisms such that $p_0 \circ i = p_1 \circ i = \textrm{id}_X$, and $\langle p_0, p_1 \rangle : P \to X \times X$ is a fibration.
As usual, $p_0, p_1 : P \to X$ are trivial fibrations, so they are isomorphisms of 0-skeleta.
Hence $i : X \to P$ is also an isomorphism of 0-skeleta.
But that means every path in $X$ is constant – in particular, there are no paths between distinct points, nor loops.
So the only objects in our subcategory are discrete sets.

The arguments above don't actually use anything special about simplices.
The key point is that a semisimplicial set has no degeneracies, so there are no semisimplicial maps $X \to Y$ if $X$ has an $n$-simplex and $Y$ does not have any $n$-simplices.
(In particular, $\Delta^0$ is not a terminal object!)
So the same argument applies to e.g. the category of cubical sets without degeneracies, or the category of globular sets without degeneracies, etc.


问:
Vector field commutator and wave equation
I am studying this review on black holes https://arxiv.org/abs/0811.0354 by Dafermos and Rodnianski, and I try  to prove the proposition E.0.1. I  am currently stuck on the following equation :
$$
{\mathcal L}_X (\nabla_a \nabla_b \psi)-\nabla_a {\mathcal L}_X
\nabla_b\psi=2 \left ((\nabla_b\, ^X\pi_{a\mu})   -
(\nabla_\mu\, ^X\pi_{b a} ) +(\nabla_a \,^X\pi_{\mu b}) \right) \nabla^\mu\psi
$$
where
$
{}^X\pi_{\mu\nu}= =\frac12 (\mathcal{L}_X g)_{\mu\nu}.
$, $\mathcal{L}_X$ the Lie derivative of the vector field $X$, $\nabla$ the covariant derivative and $\psi$ a scalar function.
Can anyone help me to prove this formula ?

答:
Proof:  $\nabla_a\nabla_b\Psi$ is a rank-2 covariant tensor so the Lie derivative is:
$$ \mathcal{L}_X\nabla_a\nabla_b\Psi=X^c\nabla_c\nabla_a\nabla_b\Psi+(\nabla_c\nabla_b\Psi)\nabla_aX^c+(\nabla_a\nabla_c\Psi)\nabla_bX^c.$$

Similarly,
$$\mathcal{L}_X\nabla_b\Psi=X^c\nabla_c\nabla_b\Psi +(\nabla_c\Psi)\nabla_bX^c.$$

So,
$$\nabla_a\mathcal{L}_X\nabla_b\Psi=(\nabla_aX^c)\nabla_c\nabla_b\Psi+X^c\nabla_a\nabla_c\nabla_b\Psi +(\nabla_a\nabla_c\Psi)\nabla_bX^c+(\nabla_c\Psi)\nabla_a\nabla_bX^c.$$

Let's concentrate on this last term. Now, taking the convention that $2\nabla_{(a}X_{b)}=\nabla_aX_b+\nabla_bX_a$, gives
$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-\nabla_a\nabla_{c}X_{b}$$

Recall the Ricci identity for a 1-form $\omega$:
$$\nabla_{a}\nabla_{b}\omega_c-\nabla_{b}\nabla_{a}\omega_c=-{R^d}_{cab}\omega_d.$$

With $\omega=X_{\flat}$ one finds:

$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-\nabla_c\nabla_{a}X_{b}+{R^d}_{bac}X_d.$$

Using $2\nabla_{(a}X_{b)}=\nabla_aX_b+\nabla_bX_a$ again gives
$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+\nabla_c\nabla_{b}X_{a}+{R^d}_{bac}X_d.$$

Using the Ricci identity again gives

$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+\nabla_b\nabla_{c}X_{a}-{R^d}_{acb}X_d+{R^d}_{bac}X_d.$$

Using $2\nabla_{(a}X_{b)}=\nabla_aX_b+\nabla_bX_a$ one last time gives

$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+2\nabla_b\nabla_{(c}X_{a)}-\nabla_b\nabla_{a}X_{c}-{R^d}_{acb}X_d+{R^d}_{bac}X_d.$$

Using the Ricci identity gives

$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+2\nabla_b\nabla_{(c}X_{a)}-\nabla_a\nabla_{b}X_{c}+{R^{d}}_{cba}X_d-{R^d}_{acb}X_d+{R^d}_{bac}X_d.$$

So we've established

$$2\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+2\nabla_b\nabla_{(c}X_{a)}+{R^{d}}_{cba}X_d-{R^d}_{acb}X_d+{R^d}_{bac}X_d.$$

Now the first (algebraic) Bianchi identity gives
$$ {R^d}_{abc}+{R^d}_{bca}+{R^d}_{cab}=0.$$

Hence,
$$2\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+2\nabla_b\nabla_{(c}X_{a)}-2{R^d}_{acb}X_d.$$

So,

$$\nabla_a\mathcal{L}_X\nabla_b\Psi=(\nabla_aX^c)\nabla_c\nabla_b\Psi+X^c\nabla_a\nabla_c\nabla_b\Psi +(\nabla_a\nabla_c\Psi)\nabla_bX^c+(\nabla_c\Psi)\Big(\nabla_a\nabla_{(b}X_{c)}-\nabla_c\nabla_{(a}X_{b)}+\nabla_b\nabla_{(c}X_{a)}-{R^d}_{acb}X_d\Big).$$

Therefore, combining and canceling terms gives

$$\mathcal{L}_X\nabla_a\nabla_b\Psi-\nabla_a\mathcal{L}_X\nabla_b\Psi=X^c\nabla_c\nabla_a\nabla_b\Psi-X^c\nabla_a\nabla_c\nabla_b\Psi -(\nabla_c\Psi)\Big(\nabla_a\nabla_{(b}X_{c)}-\nabla_c\nabla_{(a}X_{b)}+\nabla_b\nabla_{(c}X_{a)}-{R^d}_{acb}X_d\Big).$$

One final use of the Ricci identity gives on the 1-form $d\Psi$ gives

$$\mathcal{L}_X\nabla_a\nabla_b\Psi-\nabla_a\mathcal{L}_X\nabla_b\Psi=-X^c{R^d}_{bca}\nabla_d\Psi+{R^d}_{acb}X_d\nabla^c\Psi-(\nabla_c\Psi)\Big(\nabla_a\nabla_{(b}X_{c)}-\nabla_c\nabla_{(a}X_{b)}+\nabla_b\nabla_{(c}X_{a)}\Big).$$

Relabelling dummy indices and using the symmetry $R_{abcd}=R_{cdab}$ gives the result.


问:
How to find the first term common to 2 AP
Suppose there are 2 APs, AP1 and AP2 where the first terms are a1 and a2 respectively and the common difference is d1 and d2. How do you find a mathematical formula to define the AP of the common terms, i.e. their first term and the common difference, without doing random guessing

let's suppose the first term is A.P’s 3,7,11,... and the second 2,9,16,...

I know the common difference of the AP of common terms is d=LCM(d1,d2) = LCM(4,7)=28 but what about the first term.

Since the common term must have the value and are defined by their own respective APs equation

a1+(n1-1)d1= a2+(n2-1)d2

and so

3+(n1-1)4= 2+(n2-1)7

4n1-4=7n2

4(n1-1)=7n2

Now what? One thing I did was find the first n1 value divisible by 7(cause it's on the other side, LHS) and the first n2 value divisible by 4 (cause it's on the other side, the RHS). And it seems to work for me, but what if I couldn't take 4 out of the bracket like this time.

Is there are better way?

I think I need to go to Fundamental theorem of Arithmetics because I vaguely remember doing something like this

答:
The two sequences are $\langle 4n-1 \rangle$ and $\langle 7m-5 \rangle$. We solve for their equal terms.

$$4n-1=7m-5$$
$$\Rightarrow m=\frac{4(n+1)}{7}=\text{natural number}$$

Smallest $m$ is when $7$ divides $n+1$ ie, (smallest) $n=6$ yielding $m=4$. Indeed $4\cdot 6 -1 =23 = 7\cdot 4 -5$

Further values of $n$ are incremented by $7$ and that of $m$ by $4$.


问:
How to evaluate the limit of multifactorial $\lim_{n\to 0} \sqrt[n]{n!!!!\cdots !}$
It is well known that $\displaystyle \lim_{n\to \infty}\sqrt[n]{n!}=\infty$, however, if we let $n\to 0$ we have a different result with a beautiful combination of $e$ and $\gamma$, that is

$$\lim_{n\to 0}\sqrt[n]{n!}= e^{-\gamma}\tag{1}\label{1}$$

To prove result \eqref{1}, we observe that the limit attains the form of $1^{\infty}$ so we can write it as $$\lim_{n\to 0} \exp\left(\frac{\ln\Gamma(n+1)}{n}\right)\underbrace{=}_{\text{L'Hopital's rule}}\lim_{n\to 0} e^{\Gamma'(n+1)}=e^{\psi_0(1)}= e^{-\gamma}$$

Now I wish to know the limit of the following  multifactorial form  for $k\in\mathbb {Z^+}$

$$\lim_{n\to 0}\sqrt[n]{n\smash[b]{\underbrace{!! !!\cdots !}_{k}}}={?}\tag{2}\label{2}\\$$

For $k=1$ we are done above and for $k=2$ we get the limit $\sqrt{2} e^{-\frac{\gamma}{2}}$. To prove this we use the double factorial argument (see equation (5)) \begin{align}\lim_{n\to 0}\sqrt[n]{n!!}&=\lim_{n\to 0} \left(2^{\frac{n}{2}+\frac{1-\cos(\pi n)}{4}}\pi^{\frac{\cos(\pi n)-1}{4}}\Gamma\left(1+\frac{n}{2}\right)\right)^{\frac{1}{n}}\\&=\sqrt{2}\lim_{n\to 0} \sqrt[n]{\Gamma\left(1+\frac{n}{2}\right)}\\&=\sqrt 2\exp\lim_{n\to 0}\left(2^{-1} \Gamma\left(1+\frac{n}{2}\right)\psi_0\left(1+\frac{n}{2}\right)\right)\tag{L'Hopital's rule}\\&=\sqrt{2}e^{-\frac{\gamma}{2}}\end{align}

since for $k=1,2$ we have evaluated the limit. How to evaluate the limit of equation \eqref{2} for all $k>2$?

答: The main task in this problem is to construct an analytic continuation of the multifactorial function $$n!^{(k)}=n(n-k)(n-2k)\cdots$$ over the real numbers. On the set of integers, we can take advantage of modulo arithmetic to arrive at $$n!^{(k)}_i=i\cdot k^{(n-i)/k}\frac{\Gamma(1+n/k)}{\Gamma(1+i/k)}$$ which is defined whenever $n\equiv i\pmod k$. Therefore, we need to find a function which interpolates the values of $$1,\frac{k^{-1/k}}{\Gamma(1+1/k)},\frac{2k^{-2/k}}{\Gamma(1+2/k)},\cdots,\frac{(k-1)k^{-(k-1)/k}}{\Gamma(1+(k-1)/k)}$$ so we consider the ansatz $$f(x)=\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{g(x)}$$ where $g(x)\equiv1$ when $x-i\equiv0\pmod k$ and is zero otherwise. This suggests the analytic function $$\frac{\sin(\pi(x-i))}{k\sin(\pi(x-i)/k)}$$ which is unfortunately negative when $(x-i)/k$ is an odd integer. We can circumvent this by introducing a factor of $\cos(\pi(x-i)/k)$ as its sign agrees with that of the above function when $x-i\equiv0\pmod k$. It follows that we can extend the multifactorial function to the reals through $$x!^{(k)}=k^{x/k}\Gamma\left(1+\frac xk\right)\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/k}\tag1$$ It is worth noting that this is not unique, as we can multiply by a factor of $\cos(j\pi(x-i)/k)$ for some integer $j$ — here we just took $j=1$. We can now determine the limit by considering each term separately \begin{align}\lim_{x\to0}x!^{(k)/x}&=k^{1/k}\lim_{x\to0}\exp\left(\frac{\log\Gamma(1+x/k)}x\right)\lim_{x\to0}\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/kx}\\&=k^{1/k}e^{-\gamma/k}\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{h(k)}\end{align} where \begin{align}h(k)&=\lim\limits_{x\to0}\frac{\sin(\pi(x-i))\cot(\pi(x-i)/k)}{kx}=-\cot\frac{\pi i}k\lim\limits_{x\to0}\frac{\sin\pi x\cos\pi i}{kx}\\&=-(-1)^i\frac\pi k\cot\frac{\pi i}k.\end{align} Using this definition of $g(x)$, the limit evaluates to $$\lim_{x\to0}x!^{(k)/x}=\left[\frac k{e^\gamma}\prod_{i=1}^{k-1}\left(\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)^{\pi(-1)^i\cot\frac{\pi i}k}\right]^{1/k}.$$ Note that the multifactorial function in $(1)$ can be extended onto the complex plane. It is holomorphic everywhere except at negative integer multiples of $k$, similar to the gamma function.


问:
Double sequence $(x_{i,j})$ such that for every $j$ the sequence $x_{i,j}$ converges to the same limit $\bar{x}$
Consider an double sequence $(x_{i,j}) \subset \mathbb{R}^{n}$ such that for every $j$

$$ x_{i,j} \rightarrow \bar{x} \qquad \text{as} \qquad i \rightarrow \infty$$

for some $\bar{x} \in \mathbb{R}^{n}$. This means for every $j$ the sequence $(x_{i,j})$ converges to the same limit point $\bar{x}$ as $i \rightarrow \infty$.

Can I argue in the following way:

Then for every $i$ the sequence $(x_{i,j})$ admits an accumulation point denoted $x_{i}$.

It holds that $x_{i} \rightarrow \bar{x}$ as $i \rightarrow \infty$.

答:
I guess you cannot argue like that. Maybe I miscomputed, but the following should be a counterexample.

Let $x_{i,j}= \left( 1+\frac 1 i\right)^{i+j}$, then for each fixed $j \in \mathbb N$ you have
$$\lim_{i\to \infty} \left( 1+\frac 1i\right)^{i}\left( 1+\frac 1i\right)^{j}=e\cdot 1 =e$$
but for a fixed $i \in \mathbb N$ you have $$\lim_{j\to \infty} \left( 1+\frac 1i\right)^{i}\left( 1+\frac 1i\right)^{j}=\left( 1+\frac 1i\right)^{i}\cdot \lim_{j\to \infty}\left( 1+\frac 1i\right)^{j} = \infty\,$$
since $1+\frac 1i >1$.


问:
Confusion about statistical moments
Consider some probability density function $p(x)$, the $n$th-moment is given by
$$
\mathbb{E}[x^n]=\int x^np(x)dx.
$$
If one considers $\mathbb{E}\left[\sum_i x_i\right]$, it makes sense$^1$ (?) to
$$
\mathbb{E}\left[\sum_i x_i\right] =  \sum_i \int x_i p(x_i)dx_i.
$$
But when it comes to $\mathbb{E}\left[(\sum_i x_i\right)^2]$ things get trickier
$$
\mathbb{E}\left[\left(\sum_i x_i\right)^2\right] =?  \sum_{i,j} \int x_i x_j p(?)dx_?,
$$
what should this expression be like?

Simply beause it gives me the right answer...

答:
$$\mathbb E\left(\left(\sum_i x_i\right)^2\right)=\mathbb E\left(\sum_i x_i^2\right)+2\mathbb E\left(\sum_{i>j} x_ix_j\right)=\sum_i\mathbb E\left( x_i^2\right)+2\sum_{i>j}\mathbb E\left( x_ix_j\right)
\\=\sum_i\int x_i^2p(x_i)\,dx_i+2\sum_{i>j}\iint x_ix_jp_{ij}(x_i,x_j)\,dx_i\,dx_j
\\=\sum_i(\sigma_i^2+\mu_i^2)+2\sum_{i>j}(\text{cov}(x_i,x_j)+\mu_i\mu_j).$$

If the rv are independent,

$$\iint x_ix_jp_{ij}(x_i,x_j)\,dx_i\,dx_j=\iint x_ix_jp_i(x_i)p_j(x_j)\,dx_i\,dx_j
\\=\int x_ip_i(x_i)\,dx_i\int x_jp_j(x_j)\,dx_j
$$
giving the expectation
$$\sum_i(\sigma_i^2+\mu_i^2)+2\sum_{i>j}\mu_i\mu_j=\sum_i\sigma_i^2+\left(\sum_i\mu_i\right)^2.$$

If the rv are also identically distributed,

$$n\sigma^2+n^2\mu^2.$$

Title: Confusion about statistical moments


问:
Determine the homomorphisms $S_n \to \mathbb{C}^*$ for $n \geq 2$.
This is an homework question:

Determine the homomorphisms $S_n \to \mathbb{C}^*$ for $n \geq 2$.

If $f: S_n \to \mathbb{C}^*$ is an homomorphism and since $\mathbb{C}^*$ is abelian, then here exist an homomorphism:

$$h: S_n/ [S_n,S_n]=A_n \to \mathbb{C}^*$$

where $f = h \circ \varphi$. Since $S_n/ [S_n,S_n]= S_n / A_n$, then there exist an homomorphism:

$$h: S_n / A_n \to \mathbb{C}^*$$

So this implies that $h(A_n)=1$ and $h(S_n-A_n)= \pm 1$. So $f$ should suffice:  $f(\sigma)=h(\sigma A_n)=h(A_n)1$ for $\sigma \in A_n$ and $f(\sigma)=h(\sigma A_n)=h(S_n - A_n) = 1$ for $\sigma \in (S_n -A_n)$.

My questions are: a) Is this correct ? b) Are there other ways to see this ?

答: Looks quite good. Assuming that you allow the homomorphism to map elements of $S_n\setminus A_n$ either all to $+1$ or all to $-1$. This is implied in some parts of your answer but not all of them. In other words, for the sake of completeness you could explain, why both cases occur.</p>


问:
$x^4-mx^3-2mx^2+2m^2x=0$ roots in arithmetic progression
For which real values of $m$ roots of equation
$$x^4-mx^3-2mx^2+2m^2x=0$$
are in arithmetic progression?

I managed to find a solution using a lot of casework, i.e. factoring the equation into
$$x(x-m)(x^2-2m)=0$$
and then finding $m$ for which it has one root, three distinct roots and four distinct roots but I think there may be an easier solution and I'm just overcomplicating it. In addition, I'm not quite sure if I should include the one root case.

Could someone guide me to a better solution? Thanks.

答:
I assume that by "roots in arithmetic progression" you mean that the four roots are of the form
$$
\alpha,\quad\alpha+k,\quad\alpha+2k,\quad\alpha+3k
$$
for some $\alpha$ and some $k\neq0$.

Now we know that $0$ and $m$ are roots. Since the coefficient of $x^3$ is $-m$ the sum of the roots must be $m$ which gives $\alpha=-m/2$ and $k=m/2$, i.e the four roots are
$$
-\frac m2,\quad0,\quad\frac m2,\quad m.
$$
Since one of the roots is $0$ the coefficient of $x$ is the negative of the product of the three non-zero roots, i.e.
$$
2m^2=\frac{m^3}4
$$
whose only non-zero solution is $m=8$.

On the other hand, if you do not assume that the four roots are distinct, since anyway the four roots are $\{-\beta,0,\beta,m\}$ for some $\beta$ one gets two extra-cases, namely $\beta=0$ and $\beta=m$.

Working again with the linear coefficient we soon get that

$\beta=0$ occurs only for $m=0$ (the polynomal is just $x^4$);

$\beta=m$ occurs only for $m=2$.


答: NO. Suppose $x_{i,j}=j$ when $j\ge i$ and $x_{i,j}=0$ when $j<i .$ Then for every $j$ we have $\lim_{i\to\infty}x_{i,j}=0.$ But for a fixed $i$ we have $j\ge i\implies x_{i,j}=j$ so the sequence $(x_{i,j})_j$ has no accumulation point.

Title: Double sequence $(x_{i,j})$ such that for every $j$ the sequence $x_{i,j}$ converges to the same limit $\bar{x}$

Body:
Consider an double sequence $(x_{i,j}) \subset \mathbb{R}^{n}$ such that for every $j$

$$ x_{i,j} \rightarrow \bar{x} \qquad \text{as} \qquad i \rightarrow \infty$$

for some $\bar{x} \in \mathbb{R}^{n}$. This means for every $j$ the sequence $(x_{i,j})$ converges to the same limit point $\bar{x}$ as $i \rightarrow \infty$.

Can I argue in the following way:

Then for every $i$ the sequence $(x_{i,j})$ admits an accumulation point denoted $x_{i}$.

It holds that $x_{i} \rightarrow \bar{x}$ as $i \rightarrow \infty$.


问:
Proof or counterexample
Let f be a continuous function. Prove or disprove, that $f: \mathbb{R}\to\mathbb{R}$ $x\mapsto x*f(x)$ is differentiable at $0$.

I would say, this is true.  My thought: $$\lim\limits_{h \to 0}\frac{h*f(h)}{h} =\lim\limits_{h\to 0} f(h) = (f \text{ cont.})f(0)$$

答:
Your thoughts are good, but your presentation is not good !

Let $g(x):=xf(x).$ Then $g(0)=0$, thus

$$ \frac{g(h)-g(0)}{h-0}= f(h) \to f(0)$$

as $h \to 0$, since $f$ is continuous at $0$.

Title: Proof or counterexample