问:
Expression of conditional expectation using integrals, similar to expression when conditioning on an event rather than a random variable
Given an absolutely continuous random variable $(X,Y)$ with piecewise continuous density function $f(x,y)$, my goal is to show that

$E[g(X) | Y] = \frac{\int_{-\infty}^\infty g(x) f(x,Y) dx}{\int_{-\infty}^\infty f(x,Y) dx}$

for all non negative measurable functions $g: R \rightarrow R$.

Intuitively this expression makes sense, and it is very similar to when we condition on a singular event. I know that $E[g(X)] = \int^\infty_{-\infty} g(x) \int^\infty_{-\infty} f(x,y) dy dx$, and that replacing the random variable $Y$ with the event $Y = y$ yields the right hand side but with $y$ in place of $Y$. Is there a way to go from the case where we condition on events to conditioning on the random variable (or rather, the $\sigma$-algebra)?

If that is not the way to go about this, I suspect is some conditional expectation trickery which must be employed. Using the equality $E[g(x)] = E[E[g(x)|Y]]$ gets us nowhere though, and I can't think of another way to come up with an expression with a "given Y" in it.

Any suggestions would be appreciated.

答:
Since the right side is measurable w.r.t. $\sigma (Y)$ what you have to show is $\int_{Y \leq y} g(X)dP= \int_{Y\leq y}\frac{\int_{-\infty}^\infty g(x) f(x,Y) dx}{\int_{-\infty}^\infty f(x,Y) dx}dP$ for all $y$.

The left side of this equation is $$\int_{-\infty}^{y} \int_{-\infty} ^{\infty}g(x) f(x,y) dxdy$$ and the right side is $$\int_{-\infty}^{y}\frac{\int_{-\infty}^\infty g(x) f(x,y) dx}{\int_{-\infty}^\infty f(x,y)dx} f_Y(y)dy.$$  Since $\int_{-\infty}^\infty f(x,y)dx=f_Y(y)$ we see that  the two sides are equal.


问:
What does "$f''(x)$ exists" mean?
Suppose $f$ is defined in a neighborhood of x, and suppose $f''(x)$ exists.
In this situation, I know $f'$ is continuous at x, but does $f'(x)$ exist near x?

答: As $f''(x)=\displaystyle\lim_{h\to0}\frac{f'(x+h)-f'(x)}{h}$, the existence of $f''(x)$ should imply the existence of $f'$ in a neighborhood of $x$ in order to make sense $f'(x+h)$ for $|h|$ small enough.

ContentLicense: CC BY-SA 4.0

Title: What does "$f''(x)$ exists" mean?


答:
It means that for any $x$ the following limit exists

$$\lim_{h\to 0} \frac{f'(x+h)-f'(x)}{h}$$

which requires that $f'(x)$ also exists.

Title: What does "$f''(x)$ exists" mean?

Body:
Suppose $f$ is defined in a neighborhood of x, and suppose $f''(x)$ exists.
In this situation, I know $f'$ is continuous at x, but does $f'(x)$ exist near x?


答:
Since the second derivative of the function $f$ is defined as:
$$f''(x)=\frac{d}{dx}(f'(x)),$$
it's obvious that if $f''(x)$ exists in some open interval, then $f'(x)$ must exist too in that open interval.

Title: What does "$f''(x)$ exists" mean?


问:
Find a side of smaller triangle

$\triangle ABC$ has side $\overline{AB}=160 cm$, $\overline{BC}=190 cm$, and $\overline{CA}=190 cm$.
Point $D$ is along side $\overline{AB}$ and $\overline{AD} = 100 cm$. Point $E$ is along side $\overline{CA}$.

Determine the length of $\overline{AE}$ if the area of $\triangle ADE$ is three-fifths the area of $\triangle ABC$.

Is my solution correct? Thanks in advance!

答:
Suppose a triangle $T_1$ has two sides $m,n$, and that the angle between those two sides is $\theta$. If we create a second triangle $T_2$ with two sides $\alpha \cdot m, \beta \cdot n$ and keep the angle between them as $\theta$, then

Area $(T_2) \;$ = $\; \alpha \beta \;\cdot \;$ Area $(T_1)$

This can easily be seen, by first keeping one of the sides (say $n$) fixed, and multiplying the other (in this case, $\alpha \cdot m$). This intermediate triangle would have the same height as the original, but the base would get multiplied by $\alpha$, so the area would also be multiplied by $\alpha$. Then the above result follows by multiplying the other side by its factor.

In your question, side AB gets multiplied by $\frac{100}{160}=\frac{5}{8}$, whereas the area gets multiplied by $\frac{3}{5}$. Thus the second multiplier (call it $x$) must satisfy

$$\frac{5}{8} x = \frac{3}{5}$$

which yields $x=\frac{24}{25}$.

Thus the length of AE will be

$$\frac{24}{25} (190) = \boxed{182.4}$$

Title: Find a side of smaller triangle


问:
Is S a vector space?
Let S be the set of real numbers in the form (x,y). With these operations defined,

$(x,y)+(a,b)=(x, y+b)$

$\alpha(x,y)=(\alpha x,\alpha y)$

is S a vector space?

Well my answer is it can't because $(x,y)+(a,b) \neq (x+a,y+b)$, but I guess it is incomplete...
Thoughts?

答:
A vector space must verify that it's sum operator is commutative for all elements. You can easily see that in your case
$$(x,y)+(a,b)=(x, y+b) \neq (a,b+y)=(a,b)+(x,y) \text{ , for } a\neq x$$
Since $(x,y+b)\neq(a,b+y)$ in general, then your $(S,+,\cdot)$ is not a vector space.

You can also notice that your $(S,+,\cdot)$ has an infinite number of neutral elements, because
$$(x,y)+(\lambda,0)=(x,y) \phantom{a} , \phantom{a} \forall \lambda \in \mathbb{R},$$
and vector spaces verify that they must have a unique neutral element. Then we got another prove that $(S,+,\cdot)$ is not a vector space.

Title: Is S a vector space?


问:
How to solve Markov chain hitting probabilities with linear algebra
Suppose that we are given a Markov chain with three states (not necessarily irreducible) and we want to calculate the hitting probabilities for various states, say for state 1. Is it possible to solve the equations from the original $3 \times 3$ transition matrix $P$ and probability vector $a$, where $a_1 = 1, a_i = \sum_{j \in S}P(i, j)a(i), i = 2, 3$? I know that it is possible to reduce the matrix by eliminating rows corresponding to our states of interest; any constants that are caused by our states of interest can be formed by summing the columns corresponding to our states of interest, and then eliminating those columns. In our case, we would eliminate row 1 and by denoting the reduced column by $c$, our equation would be $\bar{P} \bar{a} + c = \bar{a}$, where $\bar{P}$ is the corresponding $2 \times 2$ matrix.

答: The idea occurred to me the other day, so I'll finish this thread by giving a high level description. When solving some system of equations involving Markov chain where for a function $g$, $g(x) = c(x) + \sum_{y \in S}P(x,y)g(y): x \notin A$ and $g(x) = 0, x \in A$, given a transition matrix P the task is really to solve the vector $g$ from the following equation: $Pg + C = g$, where $C$ denotes any constants that are to be added in the system. As the chain might not be able to access some states and depending on the exact configuration some values of $g$ might be computable from the get-go, such states are to be evaluated and the end result will be a linear system involving only a portion of the original components and column vectors.


问:
Bounded derivative imply convergence
Let $f$ is a $C^1$ function in $(0,\infty)$ such that $f'(x)$ is bounded in $(0,\infty)$. Does $\dfrac{f(x)}{x}$ converges when $x\rightarrow\infty$? I guess it may be have a counterexample but I have not found one.

答: No, consider for instance $$f(x)=\frac12(x+1)\left(\sin\ln(1+x)-\cos\ln(1+x)\right)=\frac{x+1}{\sqrt2}\sin\left(\ln(1+x)-\frac\pi4\right)$$

Title: Bounded derivative imply convergence


问:
A probability question on number of Balls
I have below question -

In a box, there are 2 blue and 2 red balls. I need to guess the color of the ball without replacement, I will  receive a dollar if you are correct. What is the dollar amount I should pay.

My answer is as follows -
In 1st draw, my probability of correct answer is 1/2

In 2nd draw, my probability of correct answer is 2/3

In 3rd draw, my probability of correct answer is 2/2

So, expected amount will be - 1/2+2/3+1 = 2.166667

However the correct answer is given by 2.833333.

Can you please help where I made mistake?

答:
You nailed the first two draws of (1/2), (2/3).

Then you went off the rails in two ways.

For third draw:

(1/3) chance that 1st 2 draws the same color.  When that
happens, you are certain to be right.

(2/3) chance that 1st 2 draws different colors.  When that happens
you have a (1/2) chance of being right.

Also, you overlooked that on 4th draw, you are certain to be right.

Final answer:

$$(1/2) + (2/3) + \{[(1/3) \times 1] + [(2/3) \times (1/2)\} + (1).$$


问:
How this equation converted to the other form?
I want to know how equation number 1 converted to equation number 2 ?

Equation number 1:

$$R_{TH}=\frac{R_1R_2}{R_1+R_2}$$

Equation number 2:

$$R_1=\frac{R_{TH}R_2}{R_2-R_{TH}}$$

Body:
You simply multiply, factor out and divide:

$$R_{TH} = \frac{R_1 R_2}{R_1+R_2}$$

$$(R_1+R_2)R_{TH} = R_1 R_2$$

$$R_{1}(R_{TH}-R_2) = -R_2 R_{TH}$$

$$R_1 = \frac{R_2 R_{TH}}{R_2 - R_{TH}}$$


问:
Prove that iterated $x_{i + 1} = x_i - \log x_i$ is sublinear.
Define a recurrence as: $$x_0 = n, x_{i + 1} = x_i - \log x_i, \forall i \geq 0.$$ Let $N_n$ be such that $x_{N_n} < 1$. Is it possible to show that $N_n \in o(n)$?

Some thoughts: if $x_{i + 1} = x_i - f(x_i)$ and $f$ is constant then $N_n = \frac{n}{f(1)}$ and is thus not sublinear. If $f$ is $\Omega(n)$ then $N_n$ is constant and thus trivially sublinear.  Is $f(x) = \log x$ sufficient to make $N_n$ sublinear?

Edit: As pointed out in the comments, the criteria $x_{N_n} < 1$ can never be met for this question. The correct question should be

Let $N_n$ be the smallest integer such that $x_{N_n} < 10$. Is it possible to show that $N_n \in o(n)$?

Body:
It is interesting to notice that this looks like a Newton iterative scheme for solving the equation $f(x)=a$.

Since we do not know what is $f(x)$, writing
$$x_{n+1}=x_n-\frac{f(x_n)-a}{f'(x_n)}$$ we have
$$\frac{f(x)-a}{f'(x)}=\log(x)\implies \frac{f'(x)}{f(x)-a}=\frac 1{\log(x)}$$ $$\implies\log(f(x)-a)=\text{li}(x)+c\implies
\color{blue}{f(x)=a+c\,e^{\text{li}(x)}}$$ where $\text{li}(x)$ is the logarithmic integral function.

The function $e^{\text{li}(x)}$ is, for sure, always positive and it is equal to $0$ when $x=1$ but, at this point, it is discontinuous.

Around $x=1^-$, we have
$$e^{\text{li}(x)}=-e^{\gamma } (x-1)-\frac{1}{2} e^{\gamma } (x-1)^2+O\left((x-1)^3\right)$$ while around $x=1^+$,
$$e^{\text{li}(x)}=e^{\gamma } (x-1)+\frac{1}{2} e^{\gamma } (x-1)^2+O\left((x-1)^3\right)$$

Since it is Newton, the process is quadratic.

For illustration purposes, trying for $a=123456$,  $c=\pi$ and using $x_0=30$, we have the following iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 30.000000000000000000000000000000000000000000000000 \\
 1 & \color{red}{2}6.894152524358982809170572349163498313534833470925 \\
 2 & \color{red}{2}4.325223169265492626304969903408926811220213329449 \\
 3 & \color{red}{22.}681722534270713557158529795786301254378897174602 \\
 4 & \color{red}{22.}108470739953112533158686496921157728291987229732 \\
 5 & \color{red}{22.051}702027826858474857702973272665751388334724225 \\
 6 & \color{red}{22.0511991}43990547095115344666363810856986277372042 \\
 7 & \color{red}{22.051199104963862}951102891406188800145947352942244 \\
 8 & \color{red}{22.0511991049638627160819846994044}13283687608959958 \\
 9 & \color{red}{22.051199104963862716081984699404404760615124872753}
\end{array}
\right)$$
which is typical of a second order method.


问:
How to proof that $n$ dimensional vector space $\mathbb{R}^{n}$ has n number of orthonormal basis vector?
Currently I’m studying linear algebra and encounter statement which says for 2 dimensional vector space $\mathbb{R}^{2}$ there should be 2 orthonormal basis vector which is $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$

I’m curious that is there a way to proof such statement, not only for 2 dimensional vector space but for any $n$ dimensional vector space $\mathbb{R}^{n}$, that $\mathbb{R}^{n}$ must contains n number of orthonormal basis vectors?

答:
I think your confusion arises from not having thoroughly thought through the definition of dimension. When it comes to dimension, the orthogonality or orthonormality of a basis does not matter. Any basis for an $n$-dimensional vector space contains exactly $n$ vectors. That is the definition of the dimension. The dimension of a vector space is the number of vectors in a basis for that vector space, which can be shown to be the same as

the largest possible number of linearly independent vectors in that vector space

the smallest possible number of vectors that span the vector space.


问:
GCD of $n^2-3n-1$ and $n-4$
$n$ is a natural number and after trying the division algorithm

$$\gcd(n^2-3n-1,n-4)=\gcd(n-4,n-1)=\gcd(n-1,-3)=\gcd(n-1,3)$$

For the last part I'm not sure whether it does equal to $\gcd(n-1,3)$.
If yes, then should I take the cases where $n$ is $3k+i, 0\le i\le2$?

答:
If you write as follows $$\gcd(n-4, n-1) = \gcd(n-4, 3) = \gcd(n-1, 3)$$ you will be sure i.e

$n-1 - (n-4) = 3$

$n-4 + 3 = n-1$

Yes, you can see those cases.

Title: GCD of $n^2-3n-1$ and $n-4$


问:
Confusion on Combinatorics Problem
How many ways are there to put 6 employees in 4 identical offices if at least one employee
must be assigned to each office?

So my thinking was put a person in each office and the last office gets the extra two people, therefore the number of ways is $\binom{6}{3}$.

The other set up is putting two people in the first two offices and the last two we put an extra person in there. So the number of wasys is $\binom{6}{2} \cdot \binom{4}{2}$, but this is incorrect. The right answer is: $\binom{6}{2} \cdot \binom{4}{2}\cdot\frac{1}{2}$.

Why is the $\frac{1}{2}$here?

答:
We can take it as a division of $n$ number of different objects in $m$ different groups, where some group sizes repeats.

When no size repeats then usual formula is $\dfrac{n!}{a!b!...x!}$, where $a+b+..+x=n$. It is basically total permutation divided by permutations inside each group. But when some group size repeats then we need to consider their order, suppose size $a$ repeats $\alpha$ times then we need to divide further by $\alpha !$

For this case the first possibility as you wrote is forming groups of size $(3,1,1,1)$ for which we get initially $\dfrac{6!}{3!1!1!1!}$, but here size 1 repeats 3 times so the final number should be $\dfrac{6!}{(3!1!1!1!)(3!)}$ the next possibility of forming groups is of size (2,2,1,1), In this case the final number is $\dfrac{6!}{(2!2!1!1!)(2!2!)}$. Size 2 repeated twice, size 1 also repeated twice.

Title: Confusion on Combinatorics Problem


问:
Weather forecast and probability of a bus to arrive at its destinations
A bus is making its way to a destination. If the weather conditions are favorable today, the likelihood of delay is 3%. If the weather conditions are not favorable today, the likelihood of delay is 50%. The forecast predicts that it is 20% likely that the weather conditions will be favorable today.

What is the likelihood that the bus will be delayed?

The bus has arrived, but it was delayed. Given that the bus was delayed, what is the likelihood that the weather conditions were favorable?

My answers are:

Given that P(A) = the weather condition and P(B) = the likelihood of delay

P(B|A⟯ = P(A∩B)/P(A) = 0.03/0.2 = 0.15 = 15%

(0.2 - 0.03)/0.2 = 0.85 = 85%

Are my answers correct?

答:
Chance of delay is

(1)

$$[(0.2) \times (0.03)] ~+~ [(0.8) \times (0.5)] = .406.$$

(2)

R = event of Delay

S = event of favorable conditions.

$$p(S|R) = \frac{p(\text{events}~ R~ \text{and} ~S ~\text{both occuring})}{p(R)}.$$

$$p(S|R) = \frac{0.006}{0.406}.$$


问:
Expectation of negative binomial distribution
Given $X \sim \text{NBin}(n,p)$, I've seen two different calculations for $\mathbb{E} (X)$:

\begin{align*}
&1. \mathbb{E} (X) = \frac{n}{p}, \quad \text{or}\\
&2. \mathbb{E} (Y) = \frac{n(1-p)}{p}
\end{align*}

Proof for 1.: Proof for the calculation of mean in negative binomial distribution

Proof for 2: Although I can't find a concrete proof on stackexchange, this is the expected value used in the wikipedia article for negative binomials, and I have also seen this value used in some questions here.

I've heard someone say that both are valid depending on the way you define the negative binomial, but I still don't quite understand the difference between the set-ups for the two different $\mathbb{E} (X)$.

Could someone explain their differences? Thank you!

答:
The negative binomial distribution is the sum of $n$ i.i.d. geometric distributions.

As for the Geometric, alse for the NBinomial you have 2 kinds of parametrizations

The variable counting the total trials to get $n$ successes

The variable counting the total failures to get $n$ successes

Thus you can prove your expectations in the following way:

Start from the Geometric distribution that counts how many trials you need to get the first success:

$$P(X=x)=q^{x-1}p$$

$x=1,2,3,....$ and $q=1-p$

Calculate $\mathbb{E}[X]$

$$\mathbb{E}[X]=p\sum_{x=0}^{\infty}q^{x-1}=p\sum_{x=0}^{\infty}\frac{d}{q}q^x=p \frac{d}{dq}\frac{q}{1-q}=\dots=\frac{1}{p}$$

Hence the expectation of the NBinomial counting how many trials you need to get $k$ successes is simply

$$ \bbox[5px,border:2px solid black]
{
\mathbb{E}[\Sigma_i X_i]=k\frac{1}{p}
\qquad (1)
}
$$

note that the geometric distribution counting the failures before the first success is

$$Y=X-1$$

Thus its mean is $\mathbb{E}[Y]=\frac{1}{p}-1=\frac{q}{p}$

Hence the Expectation of the NBinomial counting the number of failures before you get $k$ successes is

$$ \bbox[5px,border:2px solid black]
{
\mathbb{E}[\Sigma_i Y_i]=k\frac{q}{p}
\qquad (2)
}
$$

...that's all!


问:
Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.

Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.($a^2+b^2$, is same $b^2+a^2$), $n \in \mathbb{N}.$

I have proved the above question which appeared in one of the Math-Olympiad. And I do know the solution. Sharing the question only because the question has a cute solution.

答:
Pretty happy with approaches. Considering $n$ as a square gives one  fine way.

Consider $n=m^2=p^2+q^2$

Now, $n(n +1)= (p^2+q^2) (m^2+1)$

$=(pm+q)^2+(qm-p)^2$

Note that they are two distinct ways.

Thus, for example, $m=5k, p=4k,q=3k$

$n(n +1)= (25k^2)^2+ (5k)^2=(15k^2+4k^2)^2+(20k^2-3k^2)^2$

And we know that there are infinite numbers of the form $n=p^2+q^2$ (Pythagorean Triplets)

ContentLicense: CC BY-SA 3.0


答:
You can "flatten" the equation:

$$R_{TH}=\frac{R_1R_2}{R_1+R_2}\leftrightarrow R_{TH}R_1+R_{TH}R_2=R_1R_2$$

Then move the terms and "unflatten",

$$R_1R_2-R_{TH}R_1=R_{TH}R_2\leftrightarrow R_1=\frac{R_{TH}R_2}{R_2-R_{TH}}.$$

Title: How this equation converted to the other form?

Body:
I want to know how equation number 1 converted to equation number 2 ?

Equation number 1:

$$R_{TH}=\frac{R_1R_2}{R_1+R_2}$$

Equation number 2:

$$R_1=\frac{R_{TH}R_2}{R_2-R_{TH}}$$


问:
product of transversal is a transversal
Let $K \leq H \leq G$ where $G$ is a group, $[G:H]$ is finite and suppose that $T$ is a transversal of $K$ in $H$ and $U$ a transversal of $H$ in $G$.

Show that $TU$ is a transversal of $K$ in $G$.

Is this correct to say that for $t \in T, u \in U$ we have $uHtK$. Then knowing that $e\in H$,$uetK=utK$ and thus $TU$ is a transversal of $K$ in $G$.

答:
You need to specify if you mean right or left transversals (you are actually mixing them up). Anyway, if $T$ is a left transversal for $K$ in $H$ and $U$ is a left transversal for $H$ in $G$, then $UT$ is a left transversal for $K$ in $G$.
First of all, $G=\{utK: u \in U, t \in T\}$: take $g \in G$, then $g=uh$ for some $u \in U, h \in H$. And $h=tk$, for some $t \in T, k \in K$. Hence $g=utk$.

Uniqueness: assume $utK=u't'K$, with $u,u' \in U, t,t' \in T$, then $ut=u't'k$ for some $k \in K$ and hence $u=u'(t'kt^{-1}) \in u'H$. So $uH=u'H$, implying $u=u'$ (definition transversal!) and this implies at its turn $t=t'$ by the same argument.

Note if $U$ and $T$ would be right transversals, then $TU$ is a right transversal for $K$ in $G$.

Title: product of transversal is a transversal


问:
Let $(C, \prec)$ be the completion of $(P, <)$. Then, if $c, d \in C$, there is some $p \in P$ such that $c \prec p \prec d$.
I'm studying the book Introduction to Set Theory by Hrbacek and Jech and came across this theorem:

5.3 Theorem Let $(P, \lt )$ be a dense linearly ordered set without endpoints. Then there exists a complete linearly ordered set $(C, \prec )$ such that

(a) $P \subseteq C$.

(b) If $p,q \in P$, then $p \lt q$ if and only if $p \prec q$ ($\prec$ coincides with $\lt$ On $P$).

(c) $P$ is dense in $C$, i.e., for any $p,q \in P$ such that $p\lt q$, there is $c\in C$ with $p\prec c\prec q$.

(d) $C$ does not have endpoints.

Moreover, this complete linearly ordered set $(C, \prec )$ is unique up to isomorphism over $P$. In other words, if $(C^*, \prec ^* )$ is a complete linearly ordered set which satisfies (a)-(d), then there is an isomorphism $h$ between $(C, \prec )$ and $(C^*, \prec ^*)$ such that $h(x)=x$ for each $x \in P$. The linearly ordered set $(C, \prec )$ is called the completion of $(P, \lt )$.

Specifically I'm interested in property (c); I would like to reverse the roles of $P$ and $C$ and prove the following:

Proposition: For any $c, d \in C$ such that $c \prec d$, there is $p \in P$ with $c \prec p \prec d$.

It seems reasonable, since in practice $P$ is a model for the rationals and $C$ is a model for the real numbers, so this proposition would simply say "between any pair of real numbers there is always a rational number". In fact, the authors use this result without ever proving it a couple of lines below.

However, I am unable to prove the result. I tried the contrapositive method but got confused with the quantifiers; then I tried contradiction, but couldn't quite finish the argument.

I would like to ask for any hint that could help me prove this result, since I'm pretty much stuck here. Thank you.

答:
As the comment stated, there must be a typo in the definition of a dense subset given in condition (c). In fact, the correct definition is the following:

$P$ is dense in $C$, i.e., for any $c,d \in C$ such that $c \prec d$ there is $p \in P$ with $c \prec p \prec d$. $\tag{*}\label{*}$

Even if you accept the definition mentioned in the book, the above property ($\ref{*}$) and the uniqueness part of the theorem do not necessarily hold. For example, consider the following sets:$$A= \{ x \in \mathbb{Q} \mid x \lt 1 \} \cup \{ x \in \mathbb{Q} \mid x \gt 4 \}, \\ B= (-\infty , 1 ] \cup [2, 3] \cup [4, \infty).$$ $(A, \lt )$ is a dense linearly ordered set, $(B, \lt )$ is a complete linearly ordered set, and neither $(A, \lt )$ nor $(B, \lt )$ has endpoints. It can be easily seen that the other conditions of the theorem are satisfied. However, we can see that our required property, $\ref{*}$, does not hold; for example, for $c=2, d=3 \in B$, there is no $p \in A$ such that$$2 \lt p \lt 3.$$The uniqueness part of the theorem does not hold as well. For example, $C=(\mathbb{R}, \lt )$, as a completion of $(A, \lt )$, also satisfies the conditions of the theorem, but the isomorphism map constructed in the proof of the uniqueness cannot map $(B, \lt)$ onto $(\mathbb{R}, \lt )$ (in fact, it is an embedding).

Addendum

Please note that in topology a dense set is defined as follows.

A subset $A$ of a topological space $X$ is dense in $X$ if for every non-empty open set $U$ of $X$ we have $A \cap U \neq \varnothing$.

Now, if we apply the above definition to the order topology (without endpoints), whose open sets are arbitrary unions of the sets $\{ x \in X \mid a \lt x \lt b \}$ for any $a,b \in X$, we can conclude that

$A$ is dense in X if for every $a \lt b$ in $X$ there is some $c\in A$ such that $a \lt c \lt b$.

It is worth noting that Thomas Jech in his book Set Theory states the definition of a dense subset as follows:

A set $D \subset P$ is a dense subset if for all $a \lt b$ in $P$ there exists a $d \in D$ such that $a \lt d \lt b$.

I hope with the above reasons and evidence you conclude that such a statement in condition (c) is just a typo.


问:
Trees with vertex set
I am having hard time understanding and solving the following question:

There are exactly three trees with vertex set {1,2,3}. Note that all these trees are paths; the only difference is which vertex has degree 2. How many trees have a vertex set {1,2,3,4}?

So how I understand it is we have $K_3$ trees for the vertex set {1,2,3}. So since a tree is a path, we have a walk that repeats no vertex twice, or a path. When I drew it out, every time, only one vertex at a time has 2 degrees. (So far correct?) So what I don't understand is how we can determine the number of trees we can find for a vertex set $K_4$.

答:
There is a general formula (which is not hard to prove) for a number of spanning trees of a graph as a determinant of submatrix of its Laplace/adjacency matrix. For $K_4$ it gives

$\det \begin{pmatrix}
3 & -1 & -1 \\
-1 & 3 & -1 & \\
-1 & -1 & 3
\end{pmatrix}=16.$

Title: Trees with vertex set


问:
What is the meaning of "$H$-orbits on $G/H$?"
I am trying to solve some problem and it phrases as "...if and only if the number of $H$-orbits on $G/H$ is $2$."

I have only learned the orbit of some element, not the group itself. So if I treat $H$ as an element of some larger group(maybe G), it is implied that this larger group $G$ is acting on $G/H$. Am I correct? What is the action like? Is it necessarily a left multiplication? Can I assume that $H$ is a normal subgroup? Or can I treat $G/H$ as just a set of left cosets?

Also, is there any obvious implication that the number of this orbits is $2$?

答:
I guess $H$ is a subgroup of $G$; hence $G/H$ is the set of, say, the left cosets of $H$ (this is not necessarily a group, if $H$ is not normal in $G$). Then, $(h,gH)\mapsto h\cdot gH:=(hg)H$ is indeed an action of $H$ on $G/H$. The point-wise orbit of $gH$ is the set:

\begin{alignat}{1}
O(gH) &= \{h\cdot gH, h\in H\} \\
&=\{(hg)H, h\in H\} \\
\end{alignat}

and the point-wise stabilizer of $gH$ is the subgroup of $H$:

\begin{alignat}{1}
\operatorname{Stab}(gH) &= \{h\in H\mid h\cdot gH=gH\} \\
&= \{h\in H\mid hgH=gH\} \\
&= \{h\in H\mid g^{-1}hgH=H\} \\
&= \{h\in H\mid g^{-1}hg\in H\} \\
\end{alignat}

Now, in case $H\unlhd G$, we get $\operatorname{Stab}(gH)=H$ for every $g\in G$, and hence (Orbit-Stabilizer theorem):

$$|O(gH)||\operatorname{Stab}(gH)|=|H|, \forall g\in G \Longrightarrow |O(gH)|=1, \forall g\in G$$

So, if there are $2$ orbits only, we have in this case $|G/H|=2$.

I'm not sure if this answers any of your questions, but hopefully it helps in some way.


问:
On the representation of polynomials $\in \mathbb{K}[x_1,x_2,x_3]$
Disregarding the ordering of monomials, can a polynomial have two formal writings?

When studying symmetric polynomials my manual says that there's a unique representation of a symmetric polynomial in terms of elementary symmetric polynomials; But I think it can be generalized to all polynomials.

Let $f(x_1,x_2,x_3)$ be a polynomial $\in \mathbb{K}[x_1,x_2,x_3]$. Since $x_1 x_2,x_3$ are independent indeterminates, if there were a relation such as $x_1 = x_2^2\cdot x_1 + x_3\cdot x_1$ or the like, I could replace $x_1$ by the former relation and get a second formal writing. But if $x_1 x_2,x_3$ are independent it should not be possible. I'm not sure about this fact and I'd like some clarification.

答:
The only intrinsic way to write polynomials in $n$ variables ${x}=(x_1,...,x_n)$ is as sum
$$
P({x})=\sum_{I}a_Ix^I
$$
where $I=(k_1,...,k_n)$ is a multindex with each $k_i\geq0$ and
$x^I=x_1^{k_1}\cdots x_n^{k_n}$.

The polynomial identity principle says that two such polynomials are equal if and only if the coefficients $a_I$ of the two coincide.

For special purposes you can choose to write polynomials in some special form. For instance notice that
$$
K[x_1,...,x_n]=\left(K[x_1,...,x_{n-1}]\right)[x_n],
$$
but in the end it is just a reshuffling of the terms.


问:
Why $p_1\mid a \wedge p_2\mid a \Rightarrow p_1p_2\mid a$?
Why $p_1\mid a \wedge p_2\mid a \Rightarrow p_1p_2\mid a$ where $p_i$ is a prime number, $p_i \neq p_j$ and $a$ is a integer?

I don't fully understand that.

Also, is it true when $p_i$ is something different from a prime number?

答:
Edit

Thanks to Dietrich Burde for indicating that a counter example is the case where $p_1$ and $p_2$ are the same prime.  This answer assumes that they are distinct primes.

$p_1 | a \implies ~\exists ~r ~\text{such that} ~a = p_1 \times r.$

Therefore, $p_2 | a ~\implies ~p_2 | (p_1 \times r).$

Since $p_1$ and $p_2$ are relatively prime, this implies that

$p_2 | r \implies ~\exists ~$s$ ~\text{such that}
~r = ~p_2 \times s.$

Therefore, $p_2 \times s \times p_1 = r \times p_1 = a.$

A sufficient (but not necessarily required) condition is that $p_1$ and $p_2$ be relatively prime.  They don't have to actually be prime #'s.


问:
Compact absolute retracts have the universal extension property. Where's the mistake in this proof?
First, the definitions:

$Y$ has the universal extension property if given any normal $X,$ closed subspace $A \subseteq X,$ and continuous function $f: A \to Y,$ there exists a continuous extension of $f$ to $X.$

A normal space $Y$ is an absolute retract if for any normal $Z$ and closed subspace $Y_0 \subseteq Z$ homeomorphic to $Y,$ the space $Y_0$ is a retract of $Z.$

Here is the problem:

Show that if $Y$ has the universal extension property, $Y$ is an absolute retract. Show that if $Y$ is an absolute retract and compact, then $Y$ has the universal extension property.

Proof: The first part is easy. Given $Y_0, Z$ as above, let $g: Y_0 \to Y$ be a homeomorphism. Then $g$ has a continuous extension $f: Z \to Y$ and $g^{-1} \circ f$ is a continuous retraction. Indeed, for any $y \in Y_0,$ we have $g^{-1}(f(y)) = g^{-1}(g(y)) = y.$

For the second part, let $A \subseteq X, f : A \to Y$ be what we're given so that we wish to extend $f$ to $X.$ Since $Y$ is normal, $Y$ is completely regular, so there exists an embedding $g : Y \to [0,1]^J$ for some set $J.$ But $[0,1]^J$ has the universal property and $g \circ f : A \to [0,1]^J$ is continuous, so there exists a continuous extension $h: X \to [0,1]^J.$ Then $g^{-1} \circ h$ is the continuous extension of $f$ that we seek. Indeed, $a \in A \Rightarrow g^{-1}(h(a)) = g^{-1}(g(f(a))) = f(a).$

My proof of the 2nd part did not use the fact that $Y$ is compact, and did not use the fact $Y$ is an absolute retract in any way besides $Y$ being normal. But if it's wrong, where is the mistake?

答:
You correctly argue that $g \circ f$ has a contiuous extension $h : X \to P  = [0,1]^J$. But there is no reason why $h(X) \subset g(Y)$. Thus in general $g^{-1} \circ h$ is not well-defined.

Let us fill the gap. Since $Y$ is compact, $g(Y)$ is a closed subspace of $P$ homeomorphic to $Y$. Since $P$ is normal, there exists a retraction $r : P \to g(Y)$. Now $g^{-1} \circ r \circ h$ is the desired extension of $f$.


问:
Expected Value with probability
A fair coin is flipped 82 times. What is the expected number of times that a flip of tails is followed immediately by two consecutive flips of heads?

The answer is 10 and I got 10.25. Could someone give a try on this? Thanks

答:
A tail followed by $2$ consecutive heads - there are $80$ possible starting positions (you cannot start this sequence in 81st or 82nd toss).

So Expected Number should be $ = \frac{80}{2^3} = 10$.

I think you made a mistake of taking $82$ possible starting positions.

Title: Expected Value with probability


答:
The subgroup $H$ acts on $G/H$ by induced multiplication. Taking a equivalence class $[g]\in G/H,$ the action is defined by $h\cdot[g] = [hg]$.

Any group action on a set $X$ is an equivalence relation given by $x R y$ if and only if exists $g\in G$ such that $y=g\cdot x$. So this does partitionate $X$ into equivalence classes. In the case above, it means that there are only two equivalence classes for the action of $H$ on $G/H$.

Title: What is the meaning of "$H$-orbits on $G/H$?"

Body:
I am trying to solve some problem and it phrases as "...if and only if the number of $H$-orbits on $G/H$ is $2$."

I have only learned the orbit of some element, not the group itself. So if I treat $H$ as an element of some larger group(maybe G), it is implied that this larger group $G$ is acting on $G/H$. Am I correct? What is the action like? Is it necessarily a left multiplication? Can I assume that $H$ is a normal subgroup? Or can I treat $G/H$ as just a set of left cosets?

Also, is there any obvious implication that the number of this orbits is $2$?


问:
Conditional expected value of a branching process
Consider the theorem and proof presented here. From the $\mathbb{E}(X_{t+1})=m\mathbb{E}(X_t)$ it is easy to show that $\mathbb{E}(X_{t+1}) = m^{t+1}\mathbb{E}(X_0)$. But what still bugs me is how do we specifically show that the conditional expected value is what is claimed, namely $\mathbb{E}(X_{t} \mid X_0 = x) = xm^{t}$?

答: If the branching process starts with $x$ individuals it will evolve like $x$ i.i.d. branching processes starting with one individual. So $E(X_t|X_0=x)=xE(X_t|X_0=1)$.


问:
Fitting group of a direct product
Given the group $G=N\oplus M$, show $F(G)=F(N)\oplus F(M)$, where $F(G)$ denotes the Fitting group of $G$ (the product of all nilpotent normal subgroups).

I would say the inclusion $\supseteq$ is trivial, as $F(N)\oplus F(M)$ is itself a nilpotent normal subgroup of $G$, by characterization of the Fitting group it must be contained in $F(G)$, but the other inclusion is what worries me.

答:
For future readers:
The answer by Arturo Magidin inspired me to look at this in another way: One checks that (if I'm not mistaken) that $$0\rightarrow F(N)\rightarrow F(G)\rightarrow F(M)\rightarrow 0$$ with the canonical injection $\iota_N:F(N)\rightarrow F(G)$ and the projection $\pi_M:F(G)\rightarrow F(M)$ described by Arturo Magidin is a short exact sequence. Then $\pi_N$ is  a retract for the injection $\iota_N$, so the sequence splits and the splitting lemma gives the desired result.

Title: Fitting group of a direct product


问:
Calculate the SD of winning a lottery
$X$ People participate in lottery.

The probability function of $X$ Is $P\{X=i\} = ci^2, i=1,2,...,20$

Exactly one of the participates wins, each participant has the same probability of winning.

Someone participate the lottery each day, what is the SD of the number of times he will participate lottery until his first win?

I calculated $c$ Using $\sum_{i=1}^{20}P\{X=i\} = 1 \Rightarrow c = \frac{1}{2870}$.

Also, the probability to win a single lottery i think is $1/i$.

To calculate SD of first time winning a lottery, i thought to use geometric distribution, and by the formula for $SD$ For that distribution:

$$
SD = \sqrt{\frac{1-p}{p^2}}, p = 1/i \Rightarrow SD = \sqrt{i-1}
$$

But all the options for answer are finite values, independent of a varible, namely, not my answer.

What did i do wrong?

答: Probability to win a lottery on a given day is $p=P(\text{win})=\sum_{i=1}^{20}P(\text{win}\wedge i\text{ people participate})\\=\sum_{i=1}^{20}P(\text{win}\mid i\text{ people participate})P(i\text{ people participate})\\=\sum_{i=1}^{20}\frac1i\times ci^2=3/41$

Title: Calculate the SD of winning a lottery


答:
That there are no relations between $x_1, x_2$, and $x_3$ is part of the definition of the polynomial ring.

That there are no relations between elementary symmetric polynomials is a non-trivial theorem.

One can easily construct rings of certain polynomials where there are relations, e.g. take the ring generated by $\{x^2,y^2,xy\}$ (i.e. all polynomials without linear terms). There is a relation between these generators, namely $x^2\cdot y^2 = (xy)^2$. Note that this is a property of a certain set of generators, not the ring itself.

Title: On the representation of polynomials $\in \mathbb{K}[x_1,x_2,x_3]$

Body:
Disregarding the ordering of monomials, can a polynomial have two formal writings?

When studying symmetric polynomials my manual says that there's a unique representation of a symmetric polynomial in terms of elementary symmetric polynomials; But I think it can be generalized to all polynomials.

Let $f(x_1,x_2,x_3)$ be a polynomial $\in \mathbb{K}[x_1,x_2,x_3]$. Since $x_1 x_2,x_3$ are independent indeterminates, if there were a relation such as $x_1 = x_2^2\cdot x_1 + x_3\cdot x_1$ or the like, I could replace $x_1$ by the former relation and get a second formal writing. But if $x_1 x_2,x_3$ are independent it should not be possible. I'm not sure about this fact and I'd like some clarification.


问:
What are some reasonable things to prove about the Collatz Conjecture?
I am writing an undergraduate paper on the $3n+1$ problem, and I am looking for some theorems related to the problem that would be reasonable for someone with my mathematical background to prove. I'm near the end of my undergraduate studies. I have had a pretty basic introduction to Number Theory and Group Theory. I have also taken a few introductory Computer Science courses. Those are the only courses that I've taken that seem to be relevant to the problem but otherwise my math classes have been: the usually calculus sequence, real analysis, introductory topology, linear algebra.

A good answer to my question would do the following:

Either state one or more non-trivial theorems related to the Collatz conjecture, or reference sources that would contain this kind of information.

Would NOT state any proofs unless a proof or partial proof would be essential for someone with my requisites if they are to be able to provide a full proof.

Theorems that I have already proved:

If $L(n)$ is a function that counts the number of mappings in the Collatz function for $n$ to get to 1, then if $n = 8k+4$, $L(n) = L(n+1)$.

If $n = 128k+28$ then $L(n) = L(n+2)$.

答:
One important relation to other number theoretic problems is that of the distance of perfect powers of 2 and 3, for which a lot of material exists, even online accessible. The existence of the so-called "1-cycles"1 (which is a whole class of possible cycles of arbitrary length but a certain simple structure) is directly dependent on the question, whether $$ {k3^N-1 \over k2^N-1}=2^A $$ exists in natural numbers N,A and k (besides for "trivial" solution (N,A,k)=(1,1,1), which leads to the (unsolved) problem observed in the Waring-conjecture whether $$ (3/4)^N \lt \{ (  3/2)^N \} \lt 1-(3/4)^N $$ $ \qquad \qquad $ where $ \{ \cdot \}$ means the fractional part (see for instance mathworld)

Here is also the concept of K.Mahler's "z-numbers" relevant, which also relates the "1-cycles" in a general view to the powers of 3/2 and is also not too difficult to understand for a treatize which focuses an undergraduate publicum.

1Simons, John; de Weger, Benne, Theoretical and computational bounds for (m)-cycles of the (3n+1)-problem, Acta Arith. 117, No. 1, 51-70 (2005). eudml, ZBL1136.11019.


问:
If $f_n \to f$ in distribution then does $\int_0^1 f_n \to \int_0^1 f$?
Suppose $(f_n)_{n\geq 1}$ be a sequence of Lebesgue integrable functions on $(0,1)$. Suppose $\sup_{n\geq 1} |f_n|$ is a well-defined and measurable function. If $f_n \to f$ in distribution then does $\int_0^1 f_n \to \int_0^1 f$?

答: On $(0,1)$ with Lebesgue measure let $f_n(x)= nI_{(0,\frac1  n)}(x)$ and $f=0$. Then $sup_n |f_n(x)|=\frac  1 x <\infty$ for each $x$ and $f_n \to f$ in distribution but $\int f_n =1$ and $\int f=0$.


问:
How can I prove the existence of $X$ for the matrix equation $W^{*}XY = M$ for all $M$?
Lemma:

Suppose $W$ and $Y$ are matrices satisfying $ker(W)=0$ and $ker(Y)=0$. Then for every matrix $M$ there exists a solution $X$ to the equation

$$
W^*XY = M.
$$

Attempt:

Since $ker(W)=0$, we have a unique solution $x$ to the equation $Wx = b$ and the inverse $W^{-1}$ exists.

Similarly, the inverse $Y^{-1}$ exists. If $W^{-1}$ exists, we know $(W^*)^{-1}$ exists as well since $(WW^{-1})^* = I^*$.

Then, for any matrix $M$ (with appropriate dimension), there exists $X$ and $X = (W^*)^{-1}MY^{-1}$.

Is this proof sound? If not, how should I prove the given statement?

Edit:
There is no explicit mentioning of these matrices being square and so I think my approach is wrong. I am not sure how to prove it otherwise.

Source: page 209 from the textbook (link: http://read.pudn.com/downloads143/ebook/625700/A%20Course%20In%20Robust%20Control%20Theory.pdf)

答:
A helper statement (please check if you are already familiar with it, if not, I can fill in the gaps):

Let $M$ be an $m\times n$ matrix over any field. If $\ker(M)=0$, then there exists an $n\times m$ matrix $M'$ such that $M'M=I_n$.

Knowing $\ker(W)=0$ and $\ker Y=0$, this means that we can find such matrices $Y'$ for $Y$ (with $Y'Y=I$) and $W'$ for $W$ (with $W'W=I$). Note that $W^*(W')^*=(W'W)^*=I^*=I$. Now, for any matrix $M$ of an appropriate size, and look at this:

$$W^*((W')^*MY')Y=(W^*(W')^*)MY'Y=IMI=I$$

Thus we can take $X=(W')^*MY'$ as a solution of your equation $W^*XY=M$.


问:
How do you parametrize the paraboloid to find the flux of $F$ across $S$?
Evaluate the surface integral double integral of $S$ of $F * dS$ for the given vector field $F$ and the oriented surface $S$. In other words, find the flux of $F$ across $S$. For closed surfaces, use the positive (outward) orientation.

$$\vec{F}(x,y,z)= y\mathbf{j} - z\mathbf{k}$$
$S$ consists of the paraboloid $y = x^2 + z^2, 0 \leq y \leq 1,$ and the disk $x^2 + z^2 \leq 1, y=1$

I did $x = u\cos t, y = u^2, z =u\sin t$

So to use double integral of $$\vec{F} * dS$$

For parametrization of the paraboloid how do you find $$\vec{F} $$ to be $<0,u^2,-u\sin t>$?

答: Simply note that $F=(0,y,-z)$, $y=u^2$ and $z=-u\sin(t)$. </p>


问:
Usefulness of compact boundary proving uniqueness in gluing smooth manifolds
I have a question about the following fact which used to prove the uniqueness(upto diffeomorphism) in gluing smooth manifolds:

Let $M$ be a smooth manifold and let $f_0,f_1:[0,1]\times \partial
M\to M$ be two smooth embedding that are identity map on $\partial
 M\times\{0\}=\partial M$. If $\partial M$ is compact then there is
a diffeotopy $H:M\times [0,1]\to M$ relative to $\partial M$ from
$H_0=\text{Id}_M$ to a diffeomorphism $\Phi=H_1:M\to M$ with
$\Phi\circ f_0=f_1$.

Now, proof of this can be found in the book Differential Topology by C.T.C. Wall or Introduction to Differential Topology by Klaus Jänich and T. Bröcker.

Both books assume $\partial M$ to be compact to prove the above fact, is it necessary? I mean is there any sort of counterexample without the assumption $\partial M$ is compact?

Remark: I noticed the same assumption that $\partial M$ is compact in the setting of topological manifold, for example, in Collars and concordances of topological manifolds written by M.A. Armstrong.

答:
Suppose $(M,g)$ is a riemannian manifold and suppose $\partial M$ has a unit normal vector field, pointing inside $M$, say $\nu$. For $p\in \partial M$ fixed, there is a little interval $[0,\varepsilon(p))$ on which the geodesic $\gamma_p(t)=\exp_p(t\nu(p))$ is defined. By regularity assumptions on $(M,g)$, one can chose $p \mapsto \varepsilon(p)$ as regular as $(M,g)$. Thus one can define a map
\begin{align}
E :\bigcup_{p\in \partial M}\{p\} \times [0,\varepsilon(p)) & \to  M \\ (p,t) \mapsto \gamma_p(t)
\end{align}
Because the differential of the exponential map at $0$ is the identity map, by the inverse map theorem, $E$ is a local diffeomorphism around $(q,0)\in\bigcup_{p\in \partial M}\{p\} \times [0,\varepsilon(p)) $: there exists an open neighbourhood $U_q$ of $(q,0)$ on which $E$ is a diffeomorphism onto its image. Such a neighbourhood contains an open subset $V_q \times [0,\varepsilon'(q))$.

If moreover, $\partial M$ is compact, one can chose a finite number of $q_i \in \partial M$ such that $V_{q_i}\times [0,\varepsilon'(q_i))$ covers $\partial M$. If $\varepsilon = \min \varepsilon'(q_i)$, then $\partial M \times [0,\varepsilon)$ is a neighbourhood on which $E$ is a local diffeomorphism, and we can choose $\varepsilon$ small enough such that it is injective. Thus, we have found a uniform tubular neighbourhood of $\partial M$ in $M$ isometrically embedded $E\left(\partial M \times [0,\varepsilon)\right)$.

If $M$ is not compact, one could not be able to chose $\varepsilon$ and to isometrically embed such a tubular neighbourhood. Because topology and curvature are closely related, maybe there can exist no riemannian metric on $M$ for which this construction can work. Right now I cannot figure out an example of such a manifold $M$, but one can easily construct a particular riemannian metric on a manifold with non-compact boundary such that you cannot define a uniform tubular neighbourhood. For example, the subset of $\mathbb{R}^2$ defined by
\begin{align}
M = \left\{(x,y)\in\mathbb{R}^2 ~|~ 0 \leqslant y < e^{-x^2} \right\}
\end{align}
with the induced euclidean metric is a submanifold with boundary $\partial M =\mathbb{R}\times\{0\}$, but there exists no isometric tubular neighbourhood of the form $\mathbb{R}\times [0,\varepsilon)$. This is a bad example, because here the metric is flat and we can easily deform it... But still, it is an example.

I quickly read some parts of the chapter about tubular neighbourhoods of C.T.C Wall, and there is a systematic use of riemannian metrics for the constructions. I did not find exactly where the compactness assumption is used in the lemmas before the theorem you stated, but I am almost sure it is used the way I used it above.


问:
Probability of a ball ever being chosen from an urn

There is one blue ball in an urn. Every time (and indefinitely), 5 red balls are added and immediately 1 random ball is withdrawn without being returned. What is the probability of the blue ball ever being withdrawn?

My idea was to define $A_k$ - {The blue ball is withdrawn at the $k$ step}, and then the solution is:

$\mathbb{P}(\bigcup_{k=1}^{\infty}A_k)=\sum_{k=1}^{\infty}\mathbb{P}(A_k).$

But calculating a general $\mathbb{P}(A_k)$ does not appear to be very simple (unless I'm missing something), so I'm thinking maybe there is a better way of approaching this?

Hints would be best.

答:
HINT:

The probability that the blue ball is not withdrawn in $k$ draws

$P(R) = \frac{5}{6} \times \frac{9}{10} \times .. \frac{4k+1}{4k+2} = \prod \limits_{i=1}^k \frac{4i+1}{4i+2}$

The probability that the blue ball will get withdrawn in $k$ draws $P(B \leq k)= 1 - P(R)$


问:
Condition for $k$ so that $\lim_{n\rightarrow \infty} 1 - k(1-2^{-k})^{(n/k)} = 1$
I'm going over a proof, whose last step is the following statement:

If $ k = (1-\epsilon )\cdot \log_{2}(n)$

$$ \lim_{n\rightarrow \infty} 1-k(1-2^{-k})^{(n/k)} = 1$$

I'm trying to understand why this k ensures that the limit converges to 1.

答:
As far as I can tell, the limit is $1$, for $0<\epsilon<1$, take a look at:
https://www.wolframalpha.com/input/?i=limit+when+n+goes+to+infinity+1-0.5*log2%28n%29%281-1%2F%282%5E%280.5*log2%28n%29%29%29%29%5E%28n%2F%280.5*log2%28n%29%29%29

Lets work over the limit:
\begin{align}
lim_{n\rightarrow \infty}[1-(1-\epsilon)log_2(n)[1-\frac{1}{2^{(1-\epsilon)log_2(n)}}]^{\frac{n}{(1-\epsilon)log_2(n)}}]&=\\
lim_{n\rightarrow \infty}[1-(1-\epsilon)log_2(n)[1-\frac{1}{n^{(1-\epsilon)}})]^{n^{1-\epsilon}\frac{n^{\epsilon}}{(1-\epsilon)log_2(n)}}]&=\\
lim_{n\rightarrow \infty}[1-(1-\epsilon)log_2(n)e^{-\frac{n^{\epsilon}}{(1-\epsilon)log_2(n)}}]&=1-0
\end{align}

Note that $lim_{n\rightarrow\infty}(1-\frac{1}{n})^{n}=e^{-1}$ (this is a famous result see https://en.wikipedia.org/wiki/List_of_limits)

Since $lim_{n\rightarrow\infty}log_2(n)e^{ -\frac{n^\epsilon}{(1-\epsilon)log_2(n)} }=0$. To show that this last limit is $0$, we can use the fact that $e^x>e^y$ if $x>y$. For $n$ large enought, we have that $-log(n)>-\frac{n^\epsilon}{(1-\epsilon)log_2(n)}$, sice logarithms grow solower than linear functions. Hence:

$$0\leq log_2(n)e^{ -\frac{n^\epsilon}{(1-\epsilon)log_2(n)} }\leq log_2(n)e^{ -log(n)}$$

Computing this limit:
$$lim_{n\rightarrow \infty}\frac{ log_2(n)}{e^{ log(n) }}= lim_{n\rightarrow \infty}\frac{log_2(n)}{n}=0$$


问:
Question about derivation in Jordan algebra
Let $(G,\circ)$ be a Jordan algebra, then $\sigma:G\to G$ given by $$c\mapsto a\circ(b\circ c)-b\circ(a\circ c),\quad\forall c\in G,$$ is a derivation, where $a$ and $b$ are two fixed elements of $G$.</p>

答:
Let's now actually prove this.

The algebra in question is commutative and satifies $$(xy)x^2=x(yx^2).$$ It is easy to see (if the base field has at least three elements!) that this is equivalent to the identity $$(x,y,zw)+(z,y,wx)+(w,y,xz)=0,$$ where $(a,b,c)=(ab)c-a(bc)$ is the associator map.

If we write $R_x$ the map given by right multiplication by $x$ (and using that it is the same as the left multiplication), this identity is immediately seen to be equivalent to the identity $$R_{xy}R_z-R_xR_yR_z+R_{zy}R_{x}-R_{y(xz)}+R_{xz}R_{y}-R_zR_yR_x=0.$$ If we interchange $x$ and $y$ in this equation, and compute the difference of the two, and arrange things properly, we see that $$[R_z,[R_x,R_y]]+R_{[R_x,R_y](z)}=0$$ If we write $\sigma=[R_x,R_y]$, then this equation is $$[R_z,\sigma]+R_{\sigma(z)}=0,$$ and one can see at once that this last equation means precisely that the map $\sigma$ is a derivation.

Your map if $\sigma$ for $x=a$ and $y=b$.

P.S. I learned this from Schafer's book on non-associative algebras and the technique comes from old papers of Albert.


问:
How do I prove this statement about greatest common divider?
$$2\gcd(a, b) \le \gcd(2a,2b)$$
I'm pretty sure it has something to do with Bezout's lemma or it could use prime factorization, but unsure on how to go exactly about it. Please help!

Full Disclosure: This came up on a midterm and something like this may be on the final so I want to make sure I have it down for next time.

答:
It is the special case $\,d = \gcd(a,b)\,$ below.

$$ d\mid a,b\,\Rightarrow\, 2d\mid 2a,2b\,\overset{\!\rm\color{#c00}U}\Rightarrow\, 2d\mid \gcd(2a,2b)\Rightarrow\ 2d\le \gcd(2a,2b)\qquad$$

where we used $\rm\color{#c00}U$ = gcd Universal Property below (provable by Bezout as in the link)

$$ x\mid y,z\!\!\overset{\rm\color{#c00}U\!\!}\iff x\mid \gcd(y,z)\qquad$$

Alternatively employ $\,2\gcd(a,b) = \gcd(2a,2b)\,$ by the gcd Distributive Law.


问:
Can synthetic division be used to divide $8x^{5}-4x^{4}+6x^{3}-11x^{2}-2x+3$ by $\left(x^{2}-\frac{1}{4}\right)$?
This is a peculiar case because a difference of two squares is a factor so if I perform synthetic division with each factor "separately" ie $\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)$, then I get two distinct polynomials. Now I am not sure what to do with the two to get the final answer. Do I add, subtract, multiply, divide them?

答:
Let $f(x)$ be the given polynomial. Multiplying top and bottom by $4$, we get:

$$\frac{32x^{5}-16x^{4}+24x^{3}-44x^{2}-8x+12}{4x^2-1}$$

Now $f(1/2) = f(-1/2) = 0$, so we have no remainder, which means the quotient is a cubic polynomial, $ax^3+bx^2+cx+d$. Therefore:

$$(ax^3+bx^2+cx+d)(4x^2 - 1) = 32x^{5}-16x^{4}+24x^{3}-44x^{2}-8x+12$$

Comparing the $x^5$ term and the constant term on the RHS, we have that $4a = 32, a = 8$, and $-d = 12, d= -12$, as there is only one way to make these terms.

Now comparing the $x^4$ term, $a \cdot 0 + b \cdot 4 = -16$, so $b = -4$. With the $x$ term, $c \cdot -1 + d \cdot 0 = -8$, thus $c = 8$.

Therefore the quotient is $8x^3-4x^2+8x-12$.


问:
Affine Plane of Order 4 Picture?
I am unable to construct an Affine Plane of Order 4, I can construct an Affine plane of Order 3, and 2. But am unable to find the construction of four anywhere,

It would be greatly appreciated if someone was able to supply a picture.

There was a similar question in the past, Affine plane of order 4?, but it was not very useful. (Plus one of the links are dead)

答:
The dotted rectangle represents a wrap-around region, so green lines leaving one side come back in on the opposite side:

Intersections and Parallels

Each of the 15 green lines passes through 4 points, as does each of the 5 blue lines.  Each green line is parallel to (i.e., has no point in common with) the two that are vertically offset from it, as well as to one of the blue lines.  It is easy to verify that the center point and any other point determine a unique line, and so by symmetry the same is true for all 15 points in the dotted box.  Finally, any two lines clearly intersect in at most one point, and every point is on five lines, so this is indeed the affine plane of order 4.

Symmetries

This picture makes it look like the bottom vertex is special, but of course any of the 16 points can be placed at the bottom.  The 3-way symmetry and 5-way symmetry of the remaining 15 points is especially clear here, since the three rows or five columns can be rotated without changing the picture.  We can also see a 4-way symmetry (3 length-4 cycles and a length-2 cycle) involving 14 points: We can permute the columns, moving column $i$ to column ($2i$ mod $5$), and swap two rows, and the picture will look the same.  Reversing the columns corresponds to doing this permutation twice.

There is one final symmetry that is not very clear from this diagram, that involves 12 points.  This paragraph will describe this symmetry.
If we keep one line fixed (say the blue line on the left), then the remaining 12 points can be arranged in 12 different ways, allowing any point to be in any position (but then the other 11 points' positions are forced).
Note that these 12 points lie on three green parallels to the fixed blue line.
We can permute their 4 blue columns with any even permutation (the four points on each parallel staying within that parallel), but we must then also vertically rotate the 12 points (moving each parallel onto the next) according to the even permutation used.
If we imagine a tetrahedron whose 4 corners are associated with the 4 columns, then an even permutation of the corners corresponds to either (A) a 0° rotation or 180° rotation (2 opposite sides are each reversed), (B) a 120° rotation (a face is rotated) clockwise, or (C) a 120° rotation counter-clockwise. There is no vertical rotation of the 12 points for case (A), while cases (B) and (C) correspond to vertically rotating up or down.  If we think of the tetrahedron as having rotatable faces (like a Rubik's cube), then we can permute the 4 columns by rotating the faces, but each rotation also rotates the 12 points vertically.  To put a point where you want it, put it in the correct column and then rotate the opposite face (the one that doesn't contain the corner for that column) to put the point at the desired height.
The twelve possible orientations of the tetrahedron correspond to the twelve possible arrangements of the twelve points, and each point is clearly at a different position for each of the twelve orientations.
(As you might suspect, it is also possible to associate the 12 points with the vertices of a truncated tetrahedron so that the rotations of the solid object correspond to the ways the points can be rearranged.  Thinking of it that way is cleaner in some ways, but I find the relation to the above diagram less obvious that way.)

Comparison with Pentagram Diagram

The $5\times3$ symmetry is the same as is visible in the pentagram diagram, in which the largest pentagon is the same as the smallest, so the 3-way symmetry corresponds to cycling through the sizes, with a 180° rotation between levels.  The blue lines here correspond to the lines through the center in the pentagram diagram.

Title: Affine Plane of Order 4 Picture?


答:
We have that by $k = (1-\epsilon )\log_{2}(n)$

$$1-k(1-2^{-k})^{\frac n k}=1-(1-\epsilon )\cdot \log_{2}(n)\left(1-\frac1{n^{1-\epsilon}}\right)^{\frac n {(1-\epsilon ) \log_{2}(n)}} \to 1$$

indeed since as $x \to 0^+$ we have $\log (1-x) \le -x$

$$\left(1-\frac1{n^{1-\epsilon}}\right)^{\frac n {(1-\epsilon )\cdot \log_{2}(n)}} =e^{\frac n {(1-\epsilon )\cdot \log_{2}(n)}\log\left(1-\frac1{n^{1-\epsilon}}\right)}\le e^{-\frac {n^{\epsilon}} {(1-\epsilon ) \log_{2}(n)}}$$

and then

$$(1-\epsilon )\cdot \log_{2}(n)\left(1-\frac1{n^{1-\epsilon}}\right)^{\frac n {(1-\epsilon ) \log_{2}(n)}} \le (1-\epsilon )\log_{2}(n)e^{-\frac {n^{\epsilon}} {(1-\epsilon ) \log_{2}(n)}}=$$

$$=\frac{\frac {n^{\epsilon}} {(1-\epsilon ) \log_{2}(n)}}{e^{\frac {n^{\epsilon}} {(1-\epsilon ) \log_{2}(n)}}} \cdot \frac{\left[(1-\epsilon )\log_{2}(n)\right]^2}{n^{\epsilon}} \to 0\cdot 0=0$$

Title: Condition for $k$ so that $\lim_{n\rightarrow \infty} 1 - k(1-2^{-k})^{(n/k)} = 1$

Body:
I'm going over a proof, whose last step is the following statement:

If $ k = (1-\epsilon )\cdot \log_{2}(n)$

$$ \lim_{n\rightarrow \infty} 1-k(1-2^{-k})^{(n/k)} = 1$$

I'm trying to understand why this k ensures that the limit converges to 1.


问:
How many ways do we have to count $1$ to $100$ by adding $1$ or $2$ each time?
We count $1$ to $20$ and each time we add $1$ or $2$. For example, the numbers can be $1,2,4,6,8,10,12,14,15,17,19$ and $20$. How many ways do we have to count $1$ to $100$ by adding $1$ or $2$ each time?

答:
To generalize it, if you count from $1$ to $n$, let $f(n)$ denote the number of ways to do so (adding just $1$ or $2$ each time). The last number you add can be either a $1$ or a $2$, which means that before the last step, you have to have arrived at either $n-1$ or $n-2$. This leads to the recursion of the Fobonacci numbers:
$$f(n) = f(n-1)+f(n-2)$$
Analyzing the initial values, we find that $f(1)=1$ and $f(2)=1$, so you are looking for the $100^{th}$ Fibonacci number which should be
$$f(100)=354224848179261915075$$


问:
$H$ must contain every Sylow $p$-subgroup of $G$
Let $p$ be a prime factor of the order of a finite group $G$. If $H$ is a normal subgroup $G$ whose index is not a multiple of $p$, show that $H$ must contain every Sylow $p$-subgroup of $G$.</p>

答: Hint: If $N$ is normal in $G$, then $N$ contains every element of $G$ that has order coprime to $[G:N]$.</p>


问:
Confused about substituting constraints into objective in lagarange function
I have an optimization problem, say, to minimize the variance of a portfolio with weights $w$ and the constraint is that the expection should be equal to $z$.

$$
\begin{aligned}
\min \quad& \text{Var}[R_w] \\
\text{subject to }\quad & \mathbb E[R_w] = z
\end{aligned}
$$

The Lagrange will have the form
$$
L(w, \lambda) = \text{Var}[R_w] + \lambda(\mathbb E[R_w] - z)
$$

I saw in some textbooks that the Lagrange function is then simplified as
$$
L(w, \lambda) = \mathbb E[R_w^2] - z^2 + \lambda(\mathbb E[R_w] - z)
$$

That is where I feel confused. I understand that $\text{Var}[R_w]$ is substituted with $\mathbb E[R_w^2] - [\mathbb E[R_w]]^2$. But I am not sure why $\mathbb E[R_w]$ can be substituted by $z$. When can the constraint be substituted into the objective function?

答:
The short answer is: whenever it makes sense. The longer answer is that you must make sure you don't ignore any constraint. We can reformulate the problem as $$\min_x\; f(x) \\ s.t.\; g(x)=z$$
and since $f(x)=g(x^2)-[g(x)]^2=g(x^2)-z^2$, both Lagrangian forms are equivalent. The reason I reformulate the problem is to stress that $x$ is the the objective variable. Therefore substitutions that do not disregard any variables are allowed.


问:
The existence of a finite group with a given order and Sylow numbers?
Is there a finite group $G$ with order $6048=2^5.3^3.7$ satisfying the following three conditions:

The number $n_2(G)$ of Sylow $2$-subgroups is $189=3^3.7$;

The number $n_3(G)$ of Sylow $3$-subgroups is $112=2^4.7$;

The number $n_7(G)$ of Sylow $7$-subgroups is $288=2^5.3^2$?

I have tried to construct a subgroup $H$ of $G$ with small index, but I failed. Does someone have any methods to do this?

答: By https://groupprops.subwiki.org/wiki/Groups_of_order_6048, $G$ is a solvable group, however, $288$ is not a solvable Sylow $7$-number, hence the answer is not.


问:
Prove that $Y$ satisfies $Y(t)=X(t)+\epsilon\int_0^tX(t-s)BY(s)\,\mathrm{d}s.$

Consider $$\dot{Y}=(A+\epsilon B)Y,\quad Y(0)=I ,$$ where $A, B\in\mathbb{R}^{n \times n}$. Prove that the matrix solution $Y$satisfies the integral equation
$$
Y(t)=X(t)+\epsilon\int_0^tX(t-s)BY(s)\,\mathrm{d}s,
$$ where $X$ is the solution of the IVP
$$
\dot{X}=AX, \quad X(0)=I.
$$

I just learned the first-order linear equations. It is easy to see that
$$
Y(t)=e^{t(A+\epsilon B)}, \ X(t)=e^{tA}.
$$
Substitute into the integral equation, if $[A,B]=AB-BA=0$, the conclusion is trivial. However, I do not know how to prove the general case. Maybe the $\epsilon$ is vital, but I cannot figure out the details.

Appreciate any help!

答: Let $sI-A=P$. Then \begin{align}Y(t)-X(t)&=\mathcal L^{-1}[(sI-A-\epsilon B)^{-1}]-\mathcal L^{-1}[(sI-A)^{-1}]\\&=\mathcal L^{-1}[(P-\epsilon B)^{-1}-P^{-1}]\end{align} and convolution gives \begin{align}\epsilon\int_0^tX(t-s)BY(s)\,ds&=\epsilon\int_0^te^{(t-s)A}Be^{s(A+\epsilon B)}\,ds\\&=\epsilon\mathcal L^{-1}[\mathcal L[e^{tA}]\mathcal L[Be^{t(A+\epsilon B)}]]=\mathcal L^{-1}[P^{-1}\epsilon B(P-\epsilon B)^{-1}].\end{align} Hence it suffices to show that $$(P-\epsilon B)^{-1}-P^{-1}=P^{-1}\epsilon B(P-\epsilon B)^{-1}$$ which is true on left-multiplying by $P$ and right-multiplying by $P-\epsilon B$.


问:
GCD of $n^2-3n-1$ and $n-4$
$n$ is a natural number and after trying the division algorithm

$$\gcd(n^2-3n-1,n-4)=\gcd(n-4,n-1)=\gcd(n-1,-3)=\gcd(n-1,3)$$

For the last part I'm not sure whether it does equal to $\gcd(n-1,3)$.
If yes, then should I take the cases where $n$ is $3k+i, 0\le i\le2$?

答:
After help from the comments and @VIVID.

$$\gcd(n^2-3n-1,n-4)=\gcd(n-4,n-1)=\gcd(n-1,-3)=\gcd(n-1,3)$$

The cases of $n$ are $n=3k,n=3k+1,n=3k+2$ for a natural $k$, will be:

$n=3k \Rightarrow \gcd(3k-1,3)=1$

$n=3k+1 \Rightarrow \gcd(3k,3)=3$

$n=3k+2 \Rightarrow \gcd(3k+1,3)=1$

Title: GCD of $n^2-3n-1$ and $n-4$


问:
What is the Euclidean norm of a M × N matrices?
We define the euclidean norm as following:
$$ \lvert x \rvert= (\sum_{j=1}^{N}(\lvert x_j \rvert)^2)^{1/2}=(x.x)^{1/2} $$

Assume $A$ is a $M$x$N$ matrices is that true: $\lvert Ax \rvert=\lvert A \rvert\lvert x \rvert$ ? or how can we write $\lvert Ax \rvert$ ?

Thanks in advance for any help!

答:
There are various ways of defining the norm for a matrix. Here is the most common:

$$||A|| := \sup\limits_{||x|| = 1} ||Ax|| = \sup_x \dfrac{||Ax||}{||x||}$$

(Here, it is assumed that you choose the Euclidean $2$-norm for $||x||$, but you could very well choose another norm on your vector space, and get another norm for your matrices).

So then, you do not necessarily have $||Ax|| = ||A||\cdot||x||$, because since $||A||$ is defined as the supremum of $\dfrac{||Ax||}{||x||}$, then for arbitrary $x$, we have $\dfrac{||Ax||}{||x||} \leq ||A|| \implies ||Ax|| \leq ||A||\cdot||x||$.

Just for fun, here is another very used norm:

$$||A||_\infty := \sup_{i,j} |a_{ij}|$$Where the $a_{ij}$ are the coefficients of the matrix $A$.

Fortunately, there is a theorem that tells us that in a finite-dimensional vector space (so, for example, the space of all $M\times N$ matrices), all norms are equivalent, in the sense that if you have two norms $||\cdot||_1, ||\cdot||_2$, then there exist positive constants $c_1 < c_2$ such that for every element $x$ of your vector space, $c_1||x||_1 \leq ||x||_2 \leq c_2||x||_2$. This tells us that nothing really gets out of hand, and if a theorem holds with one norm, it very likely does for the others. Here is a nice pdf on it.


问:
Two definitions of Euler characteristics
In third chapter of "Differential Topology" book by Guillemin and Pollack ,Euler's characteristic is defined as I(∆,∆) and in the end of chapter, an intuitive proof is given to show how this is same as one calculated by using triangulations. They defined vector field on triangulated manifold which has source on vertex ,sink on face and saddle on edge .Is there any rigorous as well as generalized ( to n dim.)  proof of this theorem by using tools from book or other fairly basic tools ?</p>

答:
One way of dealing with this is to make their argument rigorous, but follow the idea closely. This is a different approach than the one presented by Jonathan Libgober.

Construct a smooth function $\rho:[0,2] \to [0,1]$ which has positive derivative on $[0,1)$ and constant equal $1$ on $[1,2]$. For instance, extend the function $1-\exp\left(1-\frac{1}{(1-x)^2}\right) : [0,1]\to [0,1]$ by $1$ on $[1,2]$.

For the standard $n$-simplex $\Delta_n = \{ x_0 + \ldots + x_n = 1, x_i \geq 0\}$, consider the function $f_{\Delta_n} = \rho(\|x\|) : \Delta_n \to [0,1]$. It's smooth and its gradient vector field (tangent to $\Delta_n$, i.e., taken within the plane $x_0 + \ldots + x_n = 1$) indeed "looks like" the one drawn by Guillemin and Pollack.

Choose a triangulation of your manifold $M$ and define the function $f:M \to [0,1]$ by requiring that the restriction to each simplex $F_i:\Delta_n \to M$ of the triangulation pulls back to $f_{\Delta_n}$.

While it's not a smooth function, on each simplex the gradient $\nabla f|_{\mathrm{im} F_i} = \nabla (f_{\Delta_n} \circ  F_i^{-1}) $ is well-defined, and agrees on the boundaries of the simplices, so you do get a continuous, piecewise-smooth vector field $\nabla f$.

Now there's a little technical but easy work to do: check that Poincare-Hopf (and, actually, the bulk of the book) works for continuous vector fields with isolated critical points (and/or $C^1$ maps between manifolds).

Once you have that, you need to show that indices of the critical points are as intended -- that the critical point on a $k$-simplex has index $(-1)^k$. While that seems obvious since each should "look like" the critical point of $-x_1^2 - \ldots - x_k^2 + x_{k+1}^2 + \ldots + x_n^2$, the "look like" part needs to be specified and proven, and I admit I have hard time coming up with an elegant proof. The key insight is that on each individual simplex, the statement is trivially true (in some "relative" sense) by direct computation, since $\nabla f_{\Delta_n}$ is proportional to $\nabla \|x\|^2$. We see that if $p$ is the critical point lying in the interior of a $k$-simplex $\iota:\Delta_k \hookrightarrow \Delta_n$, then on the sphere $S_\varepsilon \cap \iota(\Delta_k)$ the vector field is, up to appropriate scaling, the identity map. On the other hand, on the intersection of $S_\varepsilon$ with the plane perpendicular to $\iota(\Delta_k)$ and with the interior of $\Delta_n$, which is a sphere $S^{k-1} \subset S_\varepsilon \simeq S^{n-1}$, the vector field is - up to positive scaling - the antipodal map. Fragments of these spheres within various $n$-simplices $\Delta_n$ with $p$ on the boundary "patch up" to a $(n-k-1)$-dimensional (piecewise smooth) sphere within $S_\varepsilon$ on which the map is antipodal. Since the complement of $S^{k-1}$ in $S^{n-1}$ contracts to the perpendicular $S^{n-k-1}$ (think of $S^{n-1} \subset \mathbb{R}^n$; we actually have a homeomorphism $S^{n-1} \setminus S^{k-1} \simeq S^{n-k-1} \times \mathbb{R}^k$), algebraic topology tells us that the degree of the map $S^{n-1} \to S^{n-1}$ which is identity on $S^{k-1} \to S^{k-1}$ is exactly the degree of the map on the complement understood as a map between two things homotopy equivalent to $S^{n-k-1}$, which in this case is $(-1)^{n-k} = (-1)^n (-1)^k$, but I don't see a direct differential-topological argument that would prove that.

While it seems there's a sign error of $(-1)^n$, there's actually no error here -- Poincare duality does hold for closed manifolds. To prove that there's no error, just use $-f$ in place of $f$. Yes, odd-dimensional closed manifolds turn out to have zero Euler characteristic.

I do agree point (6) is not very clear, but I also do think there should be a shorter proof than what I sketched, and hopefully one that doesn't require using the notion of degree from algebraic topology.

As far as I understand, for (5), it's enough to (a) change the remainder in the Taylor formula $g(t) = g(0) + t g'(0) + t^2 r(t)$ to the Peano form, $g(t) = g(0) + t g'(0) + t R(t)$ where $R(t) \to 0$ as $t \to 0$, and (b) define the degree of a continuous (or: piecewise smooth) map $S^n \to S^n$, which can be done by e.g. smoothing the map and taking the smooth degree. Note that the local Lefschetz number was defined using $df-I$, so for that you only need $f$ to be $C^1$, indeed.


问:
Closed form of a recurrence relation using generating functions
It's been awhile since I have done this. The sequence is $\displaystyle a_n = a_{n-1} + 5~a_{n-2}$ with $a_{0}=0$ and $a_{1}=1$. </p>

<p>I found the generating function to be $\displaystyle G(x) = \frac{x}{(1-x-5x^2)}$ but I am lost on where to go from here. </p>

答:
Linear algebra also gives an answer: writing the recurrence as

$\begin{pmatrix}a_n \\ a_{n-1}\end{pmatrix}=\begin{pmatrix}1 & 5 \\ 1 & 0\end{pmatrix}\begin{pmatrix}a_{n-1} \\ a_{n-2}\end{pmatrix}$,

and using spectral decomposition to find powers of the matrix. Note, that the characteristic polynomial of the matrix equals

$\rho(\lambda)=\lambda^2-\lambda -5$.


问:
Proof verification: Every local homeomorphism is continuous
In an exercise I'm asked to prove the following:

Prove that every local homeomorphism is a continuous mapping

I was able to write a proof for this, but it turned out a title convoluted and I would like to know if there's any way of proving this more directly. This is what I did:

My proof

Let $f:(X,\tau) \to (Y,\pi)$ be a local homeomorphism.

Let $A \in \pi$. We want to prove that $f^{-1}(A) \in \tau$.

Let $K_x \in \tau$ be a neighbourhood of $x \in X$ such that $f(K_x) \in \pi$ and $f|_{K_x}:K_x \to f(K_x)$ is an homeomorphism.

So let's first divide the set $A$ into small pieces: So let's consider the sets: $A \cap f(K_x) \in Y$  for every $x \in X$. Now, let's send each piece back to the set $X$ using the function $f|_{K_x}$: $f|_{K_x}^{-1} (A \cap f(K_x))$, and now that we have each peace in the set $X$, let's put them all together and we must get $f^{-1}(A)$:

$$f^{-1}(A) = \bigcup _{x \in X} f|_{K_x}^{-1} (A \cap f(K_x)) $$

I proved that these sets are the same, But I'm not going to put the proof here so the post doesn't become too long.

Now we know that $A \cap f(K_x)$ is open in $f(K_x) \subseteq Y$. This means that $f|_{K_x}^{-1} (A \cap f(K_x))$ is open in $K_x \subseteq X$. So $\exists B_x \in \tau: f|_{K_x}^{-1} (A \cap f(K_x)) = \underbrace{B_x \cap K_x}_{\in \tau}$. So we have the following:

$$f^{-1}(A) = \bigcup _{x \in X} \underbrace{f|_{K_x}^{-1} (A \cap f(K_x))}_{\in \tau}$$

So, because $f^{-1}(A)$ is the union of open sets, it's open, making $f$ a continuous function.

Is my proof correct? Is there a more straightforward way of proving this? Thanks for the help

答: It looks correct, but I would do it as follows. Take $x\in X$; you want to prove that $f$ is continuous at $X$. Let $V$ be a neighborhood of $f(x)$ then; you want to prov that $f^{-1}(V)$ is a neighborhood of $x$. Pick a neighborhood $K$ of $x$ such that $f|_K$ is a homeomorphism from $K$ onto $f(K)$. Then, since $V\cap f(K)$ is a neighborhood of $f(x)$ in $f(K)$, $f^{-1}\bigl(V\cap f(K)\bigr)$ is a neighborhood of $x$. But then, since $f^{-1}\bigl(V\cap f(K)\bigr)\subset f^{-1}(V)$, $f^{-1}(V)$ is a neighborhood of $x$ too.


问:
What is the Op Category?
By what is, I do not mean the construction, the construction is easy. What I mean is what does the Op category really represent to us in terms of the original? Oh sure, it represents the dual... but that is circular reasoning.

For every category $\textbf{C}$ there is a mirror image category $\textbf{C}^{op}$. When we "walk" through $\textbf{C}$ we are walking through $\textbf{C}^{op}$ "contravariantly" BUT $f^{op}$ does something different for us.

So the question really is, what is $f^{op}$ actually doing to the internal structure? If I have some category $\textbf{C}$ and follow some path of arrows from a to b then in $\textbf{C}^{op}$ I have to take the path of arrows from b to a and this somehow tells me something relevant. In some sense it seems like an inverse. It is almost as if $f^{op}$ is some "generalized" inverse that takes us back from where we came. If that sorta is true then $\textbf{C}^{op}$ is sort of an "inverse" category that tracks all changes, a sort of "undo" class that allows us, in theory, to go back.

If that is true then it would not be much different than just saying that for every arrow in a category there is an op-arrow that sort of acts like a generalized inverse that in theory lets us reverse any path of arrows.

I guess this would make total since if a category that has inverse for all arrows is self-dual(hence the generalized inverses are just normal inverses and).

Anyways, I'm trying to get some type of intuitive understanding of the opposite category so that when it is talked about I can actually reason from it in a concrete way. This inverse analogy seems like it would work but it also seems that $f^{op}$ doesn't necessarily, by definition, have to have any relation to $f$, it just has to be any arrow that reverses direction.

This leads me to another problem I have, in the definition of the opposite category it seems that we can actually have many opposite categories. Only the arrow has to be reverses, but the definition doesn't say anything about it having to have any other relationship to $f$ other than exchanging the domain and codomain. Of course, it is clear that this is not true, I haven't been able to figure out way from the book.

I guess this is really ultimately asking how does $f^{op}$ really relate to $f$ rather than just saying it is the opposite?

答:
While it is true that we generally assign some semantic understanding to categories (e.g. the "category of sets", the "category of groups", and so on), formally speaking, a category is simply a collection of objects and morphisms that satisfy some laws. If you took a well-known category and presented it simply in terms of an abstract set of objects, morphisms, and the way the morphisms composed, you may have a very hard time working out what the category represented. Not every category has a meaningful semantic interpretation.

When you take the dual of the category, you're performing a formal construction: though you might be starting with a category that has some semantic interpretation, its dual, a priori, does not. That is, there is no reason to expect that the opposite morphisms may be interpreted as a meaningful concept with respect to the original morphisms.

In some cases, they can. A category being self-dual is an obvious example, as you point out. There are non-trivial examples, though, which often arise in Stone duality-type situations, where you have a contravariant correspondence between topological and algebraic structure. For instance, the category of sets is equivalent to the opposite of the category of complete atomic Boolean algebras. However, this is by no means the norm and, in general, the dual really is most easily described as "turning the arrows around".

Title: What is the Op Category?


问:
How many possible ordered triples $(x_1,x_2,x_3)$ are there such that $x_1$,$x_2$,$x_3$ are non-negative integers and $x_1+x_2+x_3=38$?
How many possible ordered triples $(x_1,x_2,x_3)$ are there such that $x_1$,$x_2$,$x_3$ are non-negative integers and $x_1+x_2+x_3=38$?

Can someone explain the question and the working? Not too sure on how to start it. The answer is $780$. Thanks

答:
It is the question about permutation with multiset. The number of arrangements of $38$ times $1$ and two $+$ symbols.

$$\frac{40!}{38!\times2!}$$


问:
continuity of a composed function
Let $X$ be a topological space and  $H_1:[0,0.5] \to X$ and $H_2:[0.5,1] \to X$ be continuous with $H_1(0.5)=H_2(0.5)$. Why is $H(x) :=\left\{\begin{array}{ll} H_1(x), & x \leq 0.5\\
         H_2(x), & x \geq 0.5\end{array}\right.$ continuous on $[0,1]$?

答:
This is a special case of the pasting lemma.

If $X=A\cup B$ with both $A,B$ closed in $X$, $f:A\to C$ and $g:B\to C$ are both continuous and they agree on $A\cap B$, then their "glueing"

$$h:X\to C$$
$$h(x)=\begin{cases}
f(x) &\text{when }x\in A \\
g(x) &\text{when }x\in B
\end{cases}$$

is well defined and continuous as well.

Indeed, in that situation if $F\subseteq C$ is closed, then $h^{-1}(F)=f^{-1}(F)\cup g^{-1}(F)$ which are both closed subsets of $A$ and $B$ respectively. And thus they are closed in $X$ and so is their union.

Title: continuity of a composed function


问:
Books on parametrization of closed curves
I would like to get more familiar with parametrization techniques for closed curves (on a 2D plane).

To be more specific:

Given a curve on a 2D plane I want to approximate it through a series (polynomial, Fourier etc) and parametrize the series. The approximation is important because it could be that the function of the curve is unknown but individual samples can be retrieved permitting a descent approximation. One example that I found, where the authors use the Fourier series to parametrize a picture of Einstein can be found here:
https://www.i-programmer.info/projects/119-graphics-and-games/5735-how-to-draw-einsteins-face-parametrically.html

I am wondering if there are other approaches for that problem.

Can you recommend any good books/souce that can work as a good starting point?

答:
If you write $x$ and $y$ as a linear function of a single parameter you get a line.
For example: $x=2t-3, y=4t+1$ is line. If you eliminate the parameter $t$ you get the cartesian equation of line as $\frac{x+3}{2}=\frac{y-1}{4}.$

Similarly, if we write $x=r \cos t, y=r \sin t$. If $t$ is a parameter and $r$ is fixed (say $r=3$), then we have a circle of radius 3. If you eliminate $t$ by squaring and adding the two equations, you get the cartesian equation of circle as $x^2+y^2=9$. If you fix $t=t_0$ and let $r$ be the parameter then it is a straight line  passing through origin as you divide the two equation and get the line $y=x \tan t_0$

$x=at^2, y=2at$, where $a$ is fixed and $t$ is parameter represents a parabola. If you eliminate $t$ you get $y^2=4ax$.

$x=t, y=1/t$ represent rectangular hyperbola $xy=1$.

$x=a \cos t, y= b\sin t$ represents ellipse $x^2/a^2+y^2/b^2=1$

$x=a sec t, y=b\tan t$ represents hyperbola $x^2/a^2-y^2/b^2=1$.

$x=ce^t, y=ce^{-t}$, represents rectangular hyperbola $xy=c^2$

So on and  so forth.


问:
Show that $\int_{X}(\vert \alpha f_n - \alpha f + \beta g_n -\beta g\vert)^p d\mu \rightarrow 0 $ for $ n \rightarrow \infty$ for measurable f and g
Let $f_n,g_n, f ,g $ be measureable functions and $f_n\rightarrow f, g_n\rightarrow g$ for $n\rightarrow \infty$.

We also assume that:
$\int_{X}(\vert  f_n -  f \vert)^p d\mu \rightarrow 0$ for $ n \rightarrow \infty$ and the same holds for $\int\vert g_n-g\vert^pd\mu$

Now I want to show that $\int_{X}(\vert \alpha f_n - \alpha f + \beta g_n -\beta g\vert^p d\mu \rightarrow 0 $ for $ n \rightarrow \infty$.

So far I've been able to achieve that:

$lim\int_X\vert\alpha f_n-\alpha f+\beta g_n-\beta g\vert^pd\mu\leq lim\int_X(\vert\alpha\vert\vert f_n-f\vert+\vert\beta\vert\vert g_n-g\vert)^p d\mu$

Now if I could only make some argument that would allow me to move the limit inside the integral sign I feel like I could solve this, but I've no information that allows me to use either monotonous convergence or dominated convergence, from what I can tell. Any help here would be deeply appreciated!

答:
First let $p \geq 1 $.

In any normed linear space $x_n \to x$ and $y_n \to y$ implies $x_n+y_n \to x+y$. $L^{p}$ is a normed linear space under the norm $\|f\|=(\
\int |f|^{p})^{1/p}$. Hence the result follows immediately. [It is a consequence of triangle in equality for this norm].

More explicitly the result follows from the inequality $(\int |\phi -\psi|^{p})^{1/p} \leq (\int |\phi |^{p})^{1/p}+(\int |\psi|^{p})^{1/p}$.

For $0<p<1$ the inequality you need is $(a+b)^{p} \leq a^{p}+b^{p}$ for all $a, b \geq 0$.


问:
$(x-a)^2(x-a-1) + p \equiv 0 \bmod{p^3}$
Let p be a prime and a belong to Z. Find all solutions to the equation
$$(x-a)^2(x-a-1) + p \equiv 0 \bmod{p^3}$$

I'm having a hard time working with this as such few variables are given. We know p is prime and a is an integer, and we are solving for x. I tried letting another variable $y=x-a$, but that leaves me with $$y^2(y-1)=p(kp^2-1)$$ which tells me that either $p|y$ or $p|(y-1)$, but not much about the possible values of x... I'm uncertain whether it is looking for a set of numbers (and whether that set would be dependent on a and p? I couldn't imagine it not be), or if the answer is no solution. Any advice would be great.

答:
Note that $p \|(x-a)^2(x-a-1)$, so $p \nmid x-a$ and $p \| x-a-1$. Write $x-a=rp+1$, where $p \nmid r$. We have $(rp+1)^2rp+p \equiv 0 \pmod{p^3}$, so $0 \equiv (rp+1)^2r+1 \equiv 2r^2p+(r+1) \pmod{p^2}$. Thus $p \mid r+1$. Write $r=sp-1$. We have $2(sp-1)^2p+sp \equiv 0\pmod{p^2}$, so $0 \equiv 2(sp-1)^2+s \equiv 2+s \pmod{p}$. Therefore we can write $s=tp-2$.

We now have $x=a+rp+1=a+(sp-1)p+1=a+((tp-2)p-1)p+1=tp^3-2p^2-p+(a+1)$, where $t \in \mathbb{Z}$. In other words, $x \equiv -2p^2-p+(a+1) \pmod{p^3}$.


问:
Can synthetic division be used to divide $8x^{5}-4x^{4}+6x^{3}-11x^{2}-2x+3$ by $\left(x^{2}-\frac{1}{4}\right)$?
This is a peculiar case because a difference of two squares is a factor so if I perform synthetic division with each factor "separately" ie $\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)$, then I get two distinct polynomials. Now I am not sure what to do with the two to get the final answer. Do I add, subtract, multiply, divide them?

答:
You simply may ‘compose’ the two synthetic divisions:
$$\begin{array}{*7{r}}
&8&-4&6&-11&-2&3 \\\times\tfrac12&&4&0&3&-4&-3\\\hline
&\color{red}8&\color{red}0&\color{red}6&\color{red}{-8}&\color{red}{-6}&\color{cyan}0 \\
\times-\tfrac12&&-4 &2&-4&6 \\\hline
& \color{red}8 &\color{red}{-4}&\color{red}8&\color{red}{-12}& \color{cyan}0
\end{array},$$
whence the quotient $\:8x^3-4x^2+8x-12$.


问:
Is it an abuse of notation to omit the leading zero in a decimal less than 1?
Is it acceptable to write $.001$ rather than $0.001$ when using decimal notation?

Are there contexts in which omitting the leading zero is acceptable, and other situations in which it is not?

答:
If I document matrices where there are systematic zeros, for instance in triangular matrices, then I even reduce the "0.0" to the single "." to help the reader to focus on the non-redundant, numerically relevant part.

$$ P=\small \begin{bmatrix}
 1 & . & . & . & . & . \\
 1 & 1 & . & . & . & . \\
 1 & 2 & 1 & . & . & . \\
 1 & 3 & 3 & 1 & . & . \\
 1 & 4 & 6 & 4 & 1 & . \\
 1 & 5 & 10 & 10 & 5 & 1
 \end{bmatrix} $$
Also in correlation-matrices, where it is understood, that entries cannot have absolute values greater than 1.0 I feel it improves readability when ".1234" is written instead of "0.1234".

$$ C= \small \left[ \begin{array} {rrrrrrr}
   1.   &  - .078&  - .135&  - .084&  - .015&    .039\\
  - .078&   1.   &  - .021&  - .020&    .020&  - .010\\
  - .135&  - .021&   1.   &    .052&    .052&  - .012\\
  - .084&  - .020&    .052&   1.   &    .063&    .115\\
  - .015&    .020&    .052&    .063&   1.   &  - .057\\
    .039&  - .010&  - .012&    .115&  - .057&   1.
     \end{array} \right] $$

As far as this two examples are concerned, I think that reduction is "acceptable" (and is also used in many instances in the literature).

But this reduction is unconvenient (and possibly in-acceptable) if there is a risk of misreading of numbers, for instance, where the decimal point might be overlooked and cannot be re-discovered (because of lack of systematic/redundant information) - and this is in my experience the usual occurrence of decimal numbers...


问:
A unique question in combinatorics and non-commutative variables.
Let $x,y$ be two variables that satisfy:

$xy=yx+1$ (they are not commutative).

Find $(xy)^2 ,(xy)^3, (yx)^2, (yx)^3$ as a linear  combination in terms of $y^jx^j$.
Then find a formula for $(xy)^n$ and $(yx)^n$.

After some calculations we get:

$(xy)^2=y^2x^2+3yx+1$.

$(xy)^3=y^3x^3+6y^2x^2+7yx+1$.

$(xy)^4=y^4x^4+10y^3x^3+25y^2x^2+15yx+1$.

$(yx)^2=y^2x^2+yx$.

$(yx)^3=y^3x^3+3y^2x^2+y$

Now, given $xy=yx+1$ we can use the binomial formula:

$(xy)^n=\sum_{k=0}^{n} {n \choose k} (yx)^k$.

Denote:

$(yx)^{n}=\sum a_{n,k} y^kx^k$.

Then,

$(yx+1)^n=(yx)(yx)^n=\sum {n\choose k} yxy^kx^k$.

After computing we get,

$(yx)^{n+1}= \sum a_{n,k} (ky^kx^k+y^{k+1}x^{k+1})$.

By comparing the coefficient of $[y^kx^k]$ in $(yx)^{n+1}$:

$a_{n,k}=ka_{n-1,k}+a_{n-1,k-1}$

Body:
I found different formulas, namely
\begin{align*}
(xy)^1 & = yx+1 \\
(xy)^2 & = y^2x^2+3yx+1\\
(yx)^3 & = y^3x^3+6y^2x^2+7yx+1\\
(yx)^4 & = y^4x^4+10y^3x^3+25y^2x^2+15yx+1\\
(yx)^5 & = y^5x^5+15y^4x^4+65y^3x^3+90y^2x^2+31yx+1
\end{align*}
if correct, can you guess a general formula?


问:
Joint density functions in Probability and statistics

Hi could anyone help me here. I am currently studying Probability and Statistics and I have just been introduced to the idea of Joint density functions. So for the question this is how far I have got:

$$f_{X,Y}(x,y)=\begin{cases}
 8xy& \text{if } 0< y < x < 1\\
0  &\text{otherwise},
\end{cases}$$

I know I need to find out firstly $(f_x|y)$ and $(f_y|x)$

With $(f_x|y)(x|y) = \frac{(f_(x,y)*(x,y)}{f_y*(y))}  = \frac{8xy}{?}$

here I'm not sure what should be on the bottom. I think maybe 8x?

and then secondly I need to find out secondly $(f_y|x)$

with $(f_y|x)(y|x) =  \frac{(f_(x,y)*(x,y)}{f_x*(x))}  = \frac{8xy}{?}$

and again I'm not quite sure what should be on the bottom. I think maybe 8y?

Any help on this would really be appreciated. Thank you.

答:
here I'm not sure what should be on the bottom. I think maybe 8x

the bivariate support is

$$0<\underbrace{y<x<1}_{X-Support}$$

Thus to derive $f_Y(y)$ you have to integrate the joint density over $X-$ support

$$f_Y(y)=\int_y^1 8xy dx=\dots=4y(1-y^2)\cdot\mathbb{1}_{(0;1)}(y)$$

Thus the conditional density is

$$f_{X|Y}(x|y)=\frac{2x}{1-y^2}\mathbb{1}_{(y;1)}(x)$$

Thus, given that $Y=y=\frac{1}{3}$ you get

$$f_{X|Y=\frac{1}{3}}(t)=\frac{9}{4} t\cdot \mathbb{1}_{(\frac{1}{3};1)}(t)$$

Concluding,

The probability to have

$$\mathbb{P}\Bigg[X>\frac{3}{4}\Bigg|Y=\frac{1}{3}\Bigg]=\int_{\frac{3}{4}}^{1}\frac{9}{4} tdt=\frac{63}{128}\approx0.49$$


问:
In $\Delta ABC, AB = AC = \sqrt{3}$ and $D$ is a point on $BC$ such that $AD = 1$ . Find $(BD\times DC)$ .

In $\Delta ABC, AB = AC = \sqrt{3}$ and $D$ is a point on $BC$ such that $AD = 1$ . Find $(BD \times DC)$ .

What I Tried: Here is a picture :-

I have no idea for this problem. From the picture if $D$ lies a bit near of $B$ then I just reflect it to get $E$ near $C$ , $\Delta ADE$ is an isosceles triangle, but there's no use to it, how would that help in the original problem?

In fact, that did not look like the case when I tried it in Geogebra. $D$ lies nearly in the middle of $BC$ and, also with a bit of approximation, I got $(BD\times DC)$ to be nearly equal to $2$ , but why?

Can anyone help me with a reason here? Thank You.

答:
Say $BC = 2x$. Say altitude from $A$ to $BC$ is $h$ and meets $BC$ at point $M$.

Given isosceles triangle,

$BM = MC = x$.

$h^2 = (\sqrt3)^2 - x^2 = 3 - x^2$

$DM = \sqrt {1- h^2} \,$

WLOG, assuming $D$ is left of $M$

$BD = BM - DM = x - \sqrt {1 - h^2}$

$CD = CM + DM = x + \sqrt {1 - h^2}$

$BD.DC = x^2 - 1 + h^2 = 2$


问:
complex numbers with harmonic conjugate
how do I solve this. please make it as simple as possible I am really bad at maths.

Let $$u(x, y) = \exp x  \cos y + \exp y  \cos x + xy$$
Show that u(x, y) is harmonic and find the harmonic conjugate of $u(x, y)$ with $v(0, 0) = 0$.

答:
The first Cauchy-Riemann equation is

$$\frac{\partial u}{ \partial x}= \frac{\partial v}{\partial y},$$

which gives us

$$\frac{\partial v}{\partial y} = e^x \cos(y) - e^y \sin(x) + y.$$

By integrating, we get

$$v = e^x \sin(y) - e^y \sin(x) + \frac{1}{2}y^2 + \phi(x).$$

Now, we can differentiate this with respect to $x$ and subsitute in the other Cauchy-Riemann equation, $\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}$.

$$e^x \sin(y) - e^y \cos(x) + \phi'(x) = -(-e^x \sin(y) + e^x \cos(x) + x)$$

$$ \phi'(x) = -x \implies \phi(x) = -\frac{1}{2}x^2 $$

Observe that we omit the constant of integration due to the initial condition $v(0,0)=0$.

In conclusion, we have

$$v(x,y) = e^x \sin(y) - e^y \sin(x) + \frac{1}{2}(y^2-x^2).$$

Title: complex numbers with harmonic conjugate


问:
Probability of a process being zero
I found proofs in books (for example Oksendal's) where there is a process $S_t$ which satisfies $\mathrm E[S_t^2]=0$ for all $t$. Based on this, a conclusion is made that

$$
\mathbb P[S_t = 0 \textrm{ for all } t \in \mathbb Q \cap [0,T]]=1
$$
for some $T>0$ and where $\mathbb Q$ is the set of rational numbers. Then, using the continuity of $S_t$ a further conclusion is made that
$$
\mathbb P[S_t = 0 \textrm{ for all } t \in [0,T]] = 1
$$

My question is, why can one conclude the first probability but not the second from $\mathbb E[S_t^2]=0$? Also, probably related, how do I show these conclusions rigorously?

答:
For a single $t$, we have $E(S_t^2)=0$ implies $P(S_t\neq 0)=0$. Now if $t_1,t_2,\ldots$ are any countable set of times, the probability that $S_{t_i}\neq 0$ for some $i$ is at most the sum of the individual probabilities, i.e. $0$.

But we cannot in general make the same conclusion for uncountable sets of times; for example if $S_t$ is the indicator function of a Poisson point process then we have $E(S_t^2=0)$ for any given $t$, yet almost surely there are infinitely many values of $t$ with $S_t=1$.

Title: Probability of a process being zero


问:
Definition of $k$-times differentiable on $S$?
I'm trying to come up with a good definition of being $k$-times differentiable on a subset $S$.

Here's the definitions I'm working with.

Let $f$ be a function from $X\subseteq\mathbb{R}$ to $\mathbb{R}$. Let $L\in\mathbb{R}$. If $x_0\in X$ is a limit point of $X$, we say that $f$ has derivative $L$ at $x_0$ iff $\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=L$, i.e. iff
$$\forall\epsilon>0,\exists\delta>0,0<|x-x_0|<\delta\implies\left\lvert\frac{f(x)-f(x_0)}{x-x_0}-L\right\rvert<\epsilon.$$

If $S$ is a subset of $X$, we say that $f$ is differentiable on $S$ iff the derivative exists at $x_0$ for all $x_0\in S$ (in particular, every point of $S$ must be a limit point of $X$). We say that $f$ is differentiable iff it is differentiable on its whole domain.

Are these definitions standard so far?

Now the definition I've come upon for "$k$-times differentiable" is as follows.

Let $f$ be a function from $X\subseteq\mathbb{R}$ to $\mathbb{R}$. We say that $f$ is $1$-times differentiable iff $f$ is differentiable, and the first derivative of $f$ is the function $f^{'}:X\to\mathbb{R}$, also denoted $f^{(1)}$, defined in the obvious way. For $k\geq 1$, we say that $f$ is $(k+1)$-times differentiable iff $f$ is $k$-times differentiable and $f^{(k)}$ is differentiable, and in this case the $(k+1)$th derivative is the function $f^{(k+1)}:= (f^{(k)})': X\to\mathbb{R}$. A function is said to be infinitely differentiable iff it is $k$-times differentiable for all $k$.

Seem fine so far? But then I thought, how would I say that a function $f$ is $k$-times differentiable on some subset of its domain? Well, since we want "$f$ is $1$-times differentiable" to mean the same as "$f$ is differentiable", we might want "$f$ is $1$-times differentiable on $S$" to mean the same as "$f$ is differentiable on $S$".
How would you define "$f$ is $k$-times differentiable on $S$"?

答:
Of course you can define "$f$ is differentiable on $S$" as you do. Higher derivatives are then a problem. For example, if $f'(x)$ exists for all $x \in S$, we are in general not able to define $f''(x)$ for $x \in S$ because we cannnot be sure that each $x \in S$ is a be a limit point of $S$. You could do it as follows:

Define $f$ to be $k$-times differentiable on $S$ if there exist subsets $S_i \subset X$, $0=1,\ldots,k$, such that

$S = S_k \subset S_{k-1} \subset \ldots S_2 \subset S_1 \subset S_0 = X$

Each $x \in S_{i+1} $ is a limit point of $S_i$, $i =1,\ldots, k$

$f'(x)$ exists for all $x \in S_1$, $f''(x)$ exists for all $x \in S_2$, etc.

But honestly I do not think this concept is very useful.


答: First of all, consider it mod $p$.  Either $x \equiv a$ or $a+1 \mod p$.  Then write $x = a + t p$ or $x = a+1 + tp$ ...</p>

Body:
Let p be a prime and a belong to Z. Find all solutions to the equation
$$(x-a)^2(x-a-1) + p \equiv 0 \bmod{p^3}$$

I'm having a hard time working with this as such few variables are given. We know p is prime and a is an integer, and we are solving for x. I tried letting another variable $y=x-a$, but that leaves me with $$y^2(y-1)=p(kp^2-1)$$ which tells me that either $p|y$ or $p|(y-1)$, but not much about the possible values of x... I'm uncertain whether it is looking for a set of numbers (and whether that set would be dependent on a and p? I couldn't imagine it not be), or if the answer is no solution. Any advice would be great.


问:
Question on Group Isomorphism Dihedral group
Let $ D_6 = \{ e, r, r^2,f,rf,r^2f \} $ be the dihedral group of order $ 6 $ (Recall that we have the  following relations: $ r^3=e, f^2=e $ and $ fr^i=f^{3-i}r $ for $ i=1,2 $).

Suppose that $ G=\{ u,v,w,x,y,z \} $ is a group with operation $ * $ where $ z $ has order $ 3 $, $ y * z=u $ and $ v * w =y $. Suppose that we have the isomorphism

$ \phi(f)=v, \phi(rf)=w $. Find $ \phi(r) $ and $ \phi(r^2) $.

My answer:

$ \phi(rf). \phi(f)=\phi(rf^2)=\phi(r). $

So, $ w * v= \phi(r) $. And, $ \phi(r^2) = \phi(r). \phi(r)=(w*v)*(w*v)=w*y*v $.

What am I doing wrong?

答: Note that $frf=r^2ff=r^2$ so $\phi(r^2)=v\ast w=y$. As $|z|=3$ this forces $\phi(r)=z$.


问:
Is $X_n=Z_1+Z_2+...+Z_n$ a stationary stochastic process if the $Z_i$ are iid?

Consider a random process $X(n) = \{X_n, n\geq 1\}$ where
$X_n=Z_1+Z_2+...+Z_n$ and the $Z_i$ are iid r,v's with zero mean and
variance $\sigma^2$. Is $X(n)$ stationary?

The answer is no.

So to tackle this problem we need to know what stationarity is. As I understand it, it is that if a process $X(t)$ is stationary, then $X(t+\tau)$ for any time $\tau$ will have the same distribution as $X(t)$.

Since $Z_n\sim N(0,\sigma^2)$ I know that $X_n\sim N(0,n\sigma^2)$ since $\text{var}(\sum_iZ_i)=\sum_i\text{var}(Z_i).$

Now I also want to show that $X_{n+\tau}\nsim N(0,n\sigma^2),$ but this is where I get stuck.

答: In a stationary process $(X_n)$ the distribution of $X_n$ does not depend on $n$. If you take $(Z_i)$ to be i.i.d. $N(0,1)$ then $X_n \sim N(0, n)$ so the distribution of $X_n$ depends on $n$. In particular the variance of $X_n$ does not remain a constant. Hence $(X_n)$ is not stationary.


问:
Prime elements in $\mathbb{Z}[\sqrt{5}]$
I need to determine which of the elements $3+2\sqrt{5}$, $9+4\sqrt{5}$ and $4-\sqrt{5}$ are prime elements in $\mathbb{Z}[\sqrt{5}]$, respectively which are associated.

My ansatz is as follows:

So let $x=3+2\sqrt{5}$ divide $ab$ for $a,b \in R$. Thus, there are $u,v \in \mathbb{Z}$, such that
$$ab=(a_1+b_1\sqrt{5})(a_2+b_2\sqrt{5})=(a_1a_2+5b_1b_2)+(a_1b_2+a_2b_1\sqrt{5})=(u+v\sqrt{5})(3+2\sqrt{5})=(3u+10v)+(3v+2u)\sqrt{5}.$$
How do I go from here in order to check whether $x|a$ or $x|b$?

答:
$3+2\sqrt{5}$ has norm $-11$ and $4-\sqrt{5}$ has norm $11$. Perhaps they are associates. Indeed
$$
\frac{3+2\sqrt{5}}{4-\sqrt{5}} = 2+\sqrt{5}
$$
and $2+\sqrt{5}$ is a unit because it has norm $-1$. In fact, $(2+\sqrt{5})(-2+\sqrt{5})=1$.

$9+4\sqrt{5}$ is a unit because it has norm $1$ and so $(9+4\sqrt{5})(9-4\sqrt{5})=1$.

It remains to decide whether $3+2\sqrt{5}$ is prime. Consider
$$
\frac{\mathbb Z[\sqrt{5}]}{\langle 3+2\sqrt{5} \rangle}
=
\frac{\mathbb Z[\sqrt{5}]}{\langle 3+2\sqrt{5},11 \rangle}
=
\frac{\mathbb Z[X]}{\langle X^2-5,3+2X,11 \rangle}
\cong
\frac{\mathbb Z_{11}[X]}{\langle X^2-5,3+2X \rangle}
=
\frac{\mathbb Z_{11}[X]}{\langle X^2-5,2(7+X) \rangle}
=
\frac{\mathbb Z_{11}[X]}{\langle X^2-5,X+7 \rangle}
=
\frac{\mathbb Z_{11}[X]}{\langle X+7 \rangle}
\cong
\mathbb Z_{11}
$$
which is a domain. Therefore, $3+2\sqrt{5}$ is prime.


答:
Draw $AH\perp BC$

$$AH^2+DH^2=1$$

$$AH^2+BH^2=AH^2+(BD+DH)^2=AH^2+BD^2+DH^2+2\times BD\times  DH=3$$

$$BD^2+2\times BD\times DH=BD(BD+2\times DH)=BD(BD+DE)=BD\times BE=BD\times DC=2$$

where $BD=EC$ and $BE=DC$

Title: In $\Delta ABC, AB = AC = \sqrt{3}$ and $D$ is a point on $BC$ such that $AD = 1$ . Find $(BD\times DC)$ .

Body:
In $\Delta ABC, AB = AC = \sqrt{3}$ and $D$ is a point on $BC$ such that $AD = 1$ . Find $(BD \times DC)$ .

What I Tried: Here is a picture :-

I have no idea for this problem. From the picture if $D$ lies a bit near of $B$ then I just reflect it to get $E$ near $C$ , $\Delta ADE$ is an isosceles triangle, but there's no use to it, how would that help in the original problem?

In fact, that did not look like the case when I tried it in Geogebra. $D$ lies nearly in the middle of $BC$ and, also with a bit of approximation, I got $(BD\times DC)$ to be nearly equal to $2$ , but why?

Can anyone help me with a reason here? Thank You.


问:
Find the expected value of $X$ and $Y$
$$f_{X,Y}(x,y)=\begin{cases}\frac1{\pi(r_2^2-r_1^2)}&,\text{ if }r_1^2\le x^2+y^2\le r_2^2 \\ 0 &, \text{ otherwise }\end{cases}$$

I know that the events expressed by $X$ and $Y$ are dependent since values of $X$ restrict $Y$ and vice versa. Supposedly the covariance however is $0$. But why is it that $\operatorname{E}[X]=\operatorname{E}[Y]$?

答: $X$ and $Y$ are jointly uniform on an annulus centered at the origin with inner radius $r_1$ and outer radius $r_2$.  As such, the centroid of this region is the origin and $\operatorname{E}[X] = \operatorname{E}[Y]$.  The covariance is also zero because the density is rotationally invariant about the origin.  I leave the proof of these facts as an exercise.

Title: Find the expected value of $X$ and $Y$


问:
Let $\mathbb{E}(X_i)=\mu<\infty$ for all $i>0$ and $S_n=\sum_{i=1}^nX_i$. Is $\mathbb{E}(|X_{n}S_n+n\mu^2-\mu S_n-n\mu X_n|)$ bounded?
The $X_i$'s are independent.

It is my intention to show that this is (hopefully) bounded to aid with a reuslt, however to me it doesn't look to be the case. I applied a triangle inequality to get that the above expectation is bounded by:

$$\mathbb{E}|Y_nS_n| + n|\mu^2|+\mathbb{E}|\mu S_n| + n\mathbb{E}|\mu Y_n|$$

But the 1st, 3rd, 4th terms don't look bounded to me due to dependence on $n$. Am I missing something or is the expectation simply not bounded?

答: Let $(X_i)$ be i.i.d taking values $\pm 1$ with probability $\frac  1 2$ each. Then the question is one about boundedness of $(E|S_n|)$. There are many ways of showing that this is not true.


问:
Find $x$ such that $ 2^{16^x} = 16^{2^x} $.
Find $x$ such that $ 2^{16^x} = 16^{2^x}.$

I am a bit confused, what will happen when we expand $16^{2^x}$, will we get  $4^{2^{2^x}}$ or $4^{2^{x+1}}$ ?

答: Take log on base 2, then $$16^x \log_2 2=2^x \log_2 16. \implies 16^x= 4. 2^x$ \implies 2^{4x}=2^{x+2} \implies 4x=x+2 \implies x=2/3.$$