问:
A problem on module homomorphism
Let $R$ be a commutative ring. Prove that $\text{Hom}_R(R,M)$ and $M$ are isomorphic as left $R$-mod.
Question Let $R$ be a commutative ring. Prove that $\text{Hom}_R(R,M)$ and $M$ are isomorphic as left $R$-modules.

Attempt

Define a map $\phi:\text{Hom}_R(R,M)\to M$ by $\phi(f)=f(1), \forall f\in \text{Hom}_R(R,M) $. This makes sense as any $R$-module map $R\to M$ is determined by where it sends $1$.

Define a map $\psi:M\to \text{Hom}_R(R,M)$ by $\psi(m)=f_m$, makes sense as $f_m:R\to M$ $r\mapsto rm$ is a well defined homomorphism.

$$\text{Hom}_R(R,M)\longrightarrow^\phi M\longrightarrow^\psi \text{Hom}_R(R,M) .$$
$\phi o \psi(m)=\phi(\psi(m))=\phi(f_m)=f_m(1)=m=\text{id}_M(m).$

$$M\longrightarrow^\psi \text{Hom}_R(R,M)\longrightarrow^\phi M.$$
$\psi\phi(f)=\psi(f(1))=f_{f(1)}=f=\text{id}_{\text{Hom}_R(R,M)}(f).$

I just need a clarification whether this solution is correct. I don't need alternative solution I already done.

答: $$\psi\phi(f) = \psi(f(1)) = f_{f(1)} \stackrel{!}= f = \text{id}_{\text{Hom}_R(R,M)}(f)$$ where $(!)$ is true because $(\forall r \in R) \ f_{f(1)}(r) = rf(1) = f(r)$.

Title: A problem on module homomorphism


问:
Consistency of Adams-Multon Two step method

Show that  Adams-Multon two-step method is consistent of order 4.
Where the method is defined by:
$$u_0= a,~~ u_1=b,~~
 (*)~~~  u_{i+1}=u_i+\frac{h}{12}(5f(x_{i+1},u_{i+1})+8f(x_i,u_i)-f(x_{i-1},u_{i-1})).$$

Definition:
For all $(x,u)\in G$ define $\eta=\eta(\xi)$ to be the single solution for:
$\eta'=f(\xi,\eta)$ , $\eta(x)=u$ for the initial values $(x,u)$,  then:
$\delta(x,u;h)=1/h[\eta(x+rh)+\sum_{m=0}^{r-1} a_m \eta(x+mh)]-\rho(x,\eta(x),...,\eta(x+(r-1)h);h)$ is called the local discriminatory error.The multi-step method is called consistent if $lim_{h\to 0} \delta(x,u;h)=0$
uniformly for all $(x,u)\in G$ and is consistent of order p if $|\delta(x,u;h)|\leq kh^p$ for all $(x,u)\in G$ and $h>0$ where k is a constant.

That's what I did:
We know that $u_i$ is the estimated/approximated solution of the real solution $u(x_i)$. And $f(x,u)=u'(x)$.
Using Taylor expansion of $u$ about $x_i$ we get
$$
u(x_{i+1}) = u(x_i)+ hu'(x_i)+h^2/2! u''(x_i)+ h^3/3! u'''(x_i)+h^4/4! u''''(x_i)
$$
This gives:
$$
1) u_{i+1}=u_i+hu_i'+ h^2/2 u_i''+h^3/3! u_i''' +h^4/4! u_i''''+...
$$
On the other hand,

$u_{i+1}=u_i+h/12 [(5u_i '+ 5hu_i ''+5h^2/2u_i'''+ 5h^3/3! u_i ''''+...)+8u_i '- (u_i '-hu_i'' +h^2/2! u_i'''-h^3/3! u_i''''+...)]+O(h^4)$.

Then if we compare between coefficients in equations 1 and 2 we get:

$O(h^4)=u''''(c) (-h^4)/24$.
Tyis is the discriminatory local error and it satisfies that $|O(h^4)| \leq K*h^4$ for a constant K, I am not sure why there exists such a constant!. so by definition, we get that the method described is consistent of order 4.
If we use the definition instead to show this, it shoud lead to the same result.

Is what i did enough accurate?

答:
You computed that
$$
δ_{AM3}(x,u;h)=\frac{u(x+h)-u(x)}{h}-\frac1{12}(5u'(x+h)+8u'(x)-u'(x-h))=-\frac{h^3}{24}u^{(4)}(c)=O(h^3),
$$
which fits the condition for consistency order 3. In contrast, the order 4 method would have the error computed as
$$
δ_{AM4}(x,u;h)=\frac{u(x+h)-u(x)}{h}-\frac1{24}(9u'(x+h)+19u'(x)-5u'(x-h)+u(x-2h))=O(h^4).
$$
there is one more term involved with a different coefficient sequence.

Note that the definition of the consistency error has a factor $\frac1h$ in the leading term for the difference quotient, this lowers the degree in $h$ of the error you computed by one.

For the bound of the higher order derivative note that the derivatives of the exact solution can be computed starting with $u'=f(x,y)$ as
\begin{align}
u''(x)&=f^{[1]}(x,u)=f_x+f_uu'=f_x+f_uf
\\
u'''(x)&=f^{[2]}(x,u)=f^{[1]}_x+f^{[1]}_uu'=f_{xx}+2f_{xu}f+f_uf_x+f_{uu}[f,f]+f_u^2f
\\
u^{(4)}(x)&=f^{[3]}(x,u)=f^{[2]}_x+f^{[2]}_uf=...
\end{align}
This means that you can express $u^{(4)}$ and any other derivative as a polynomial expression in $f$ and its derivatives. Then take a sufficiently large tube around the exact solution to contain all the local exact solutions with IC $u(x_i)=u_i$ used in the construction. You get the standard situation of a continuous function on a bounded/compact set, there is a maximum (of the norm).

You need that $f$ is continuously differentiable to a sufficient order, the 3rd here. This smoothness is assumed in the definition of the error order of the method. If the $f$ in a given problem is not that smooth, then the actual numerical error will have a lower order.

ContentLicense: CC BY-SA 4.0


问:
Regular Expressions Theory Questions
I have some old exercises of regular expressions without solutions and i would like to check if i understand the theory correctly. In the methodology section i show my answers/steps so far.

After the arrow ( -> ) is my answer to every problem.

1.Give the smallest possible string that languages i) and ii) don't accept

i)b*(ab)*a* -> aab

ii)b*(a+ba)*b* -> abba

2.Given the regular expressions r=a*+b* and s=ab* + ba* + b*a + (a*b)* . Find the smallest possible string that can be expressed in:

i)r but not in s -> {empty-ε}

ii)s but not in r -> aba

iii)both r and s -> {x} there is no answer

iv)cannot be expressed in any of them -> aaba

3.Give the expression that accepts strings that:

i)have 3*x a's character -> b*(b*ab*ab*a)*b*

ii)start or end with ab/ba -> (ab+ba)(a+b)* +(b+a)*(ab+ba)

iii)dont have aaa in them -> (b+ab+aab)*(a+aa+ε)

iv)every b is followed by aaa -> a*(baaa)*a*

METHODOLOGY

i) After two a the regex can't go back and accept a b.

ii) After abb cannot accept anything other than b

2)
i)The empty string because s accepts at least the characters aba

ii)aba cannot be accepted by r since it can't have a after b*

iii) No answer since the smallest possible strings aren't the same.

iv) aaba in r you can't have aab_ with _ = other than b and s can't accept aaba.

3)
i)There are always 3*x a's in the string with any number of b's.

ii)Any string between start/end is arbitary.

iii)Every possible string that does not have aaa (so states end with b).

iv)No criterion for a's but every b should be followed by aaa.

答:
There are a few errors.

2(i) isn’t right: the $(a^*b)^*$ term in $s$ accepts the empty string. However, $s$ does not accept $aa$, which is accepted by $r$.

2(iii) isn’t right: the empty string can be expressed in both.

Another possibility for 2(iv) is $bbaa$.

3(i) is correct but can be simplified to $(b^*ab^*ab^*a)^*b^*$, and $(b^*ab^*ab^*ab^*)^*$ also works.

3(iv) isn’t right: it doesn’t recognize $baaaabaaa$, for instance. $a^*(baaaa^*)^*$ will work.

Title: Regular Expressions Theory Questions


问:
hint with Bayes rule problem
The pirate Captain Queequeg has a lazy crew and suspects they are planning to stage a mutiny. Captain Queequeg's solution is to have every member of the crew roll Queequeg's lucky die. If the roll is even, the crew member must walk the plank. If the roll is odd, the crew member must take a swig of truth serum and then reveal his or her true intent. If the crew member was planning on mutiny, he or she must walk the plank, and if not, is allowed to remain on the ship. Queequeg's die has been cursed so that evens are rolled 61% of the time. Crew members are either lazy or mutinous but not both. Of the crew members who take the serum, 18% are forced to reveal they were planning mutiny.

1) What is the chance a crew member will walk the plank, given he or she is mutinous?

2) What is the chance a crew member will walk the plank, given he or she is just plain lazy (but not mutinous)?

3) What is the chance a crew member is mutinous, given he or she walked the plank?

This is what I have so far:

The probability of walking the plank seems to be

$P(\text{plank}) = P(\text{mutinous} \cap \text{odd})+P(\text{lazy} \cap \text{even})$

$P(\text{plank}) = P(\text{mutinous})P(\text{odd})+P(\text{lazy})P(\text{even})$ Rolling the die is independent of being mutinous or lazy.

$P(\text{lazy})= P(\lnot \text{mutinous})$

1) I tried applying Bayes rule, but then I need the posterior probability from part 3) first.

$$P(\text{plank} \mid \text{mutinous}) = {P(\text{mutinous} \mid \text{plank})P(\text{plank}) \over P(mutinous)}$$

Expanding the likelihood using total probability

$$P(\text{plank} \mid \text{mutinous}) = {P(\text{mutinous} \mid \text{plank})P(\text{plank}) \over P(\text{mutinous} \mid \text{plank})P(\text{plank}) + P(\text{mutinous} \mid \lnot \text{plank})P(\lnot \text{plank})}$$

Using conditional probability, substituting the joint probability for the posterior and prior

$$P(\text{plank}\mid\text{mutinous}) = {P(\text{mutinous} \cap \text{plank}) \over P(\text{mutinous}\mid\text{plank})P(\text{plank}) + P(\text{mutinous}\mid\lnot\text{plank})P(\lnot\text{plank})}$$

Being mutinous and walking the plank aren't independent events, so I am not sure where to go from here.

答:
1)

Let us Consider how a mutinous crew member might walk the plank

he rolls an even whose probability = 0.61

he rolls an odd and is forced to reveal about mutiny = 0.39 * 0.18 = 0.0702

thus total prob = 0.61 + 0.0702 = .61702

2)

Let us Consider how a lazy crew member might walk the plank

he rolls an even whose probability = 0.61

he rolls an odd and is forced to reveal about mutiny = 0.39 * 0 = 0.0

Since he did not plan the mutiny how can he reveal it therefore this is 0

thus total prob = 0.61

3)

let us consider how a random crew member might walk the plank

P(Plank) = P(Plank|Mutinous) * P(Mutinous) + P(Plank|Lazy) * P(Lazy)

P(Plank) = (0.61702 + 0.61) * 1/2

P(Plank) = 0.61351

P(Mutinous|Plank) = (P(Plank|Mutinous) * P(Mutinous)) / P(Plank)

P(Mutinous|Plank) = 0.502

ContentLicense: CC BY-SA 3.0

Title: hint with Bayes rule problem


问:
Determine if the sequence $\{a_n\}:a_n=(-3)^n$ is an increasing or a decreasing sequence
Determine if the sequence $\{a_n\}:a_n=(-3)^n$ is an increasing or a decreasing sequence.

The $(n+1)^{st}$ term is $a_{n+1}=(-3)^{n+1}.$ So $\dfrac{a_{n+1}}{a_n}=\dfrac{(-3)^{n+1}}{(-3)^n}=\dfrac{(-3)^n(-3)^1}{(-3)^n}=-3<1.$ Doesn't this mean that $a_{n+1}<a_n$ for every $n$ which obviously isn't true? The sequence looks like $-3,9,-27,81...,$ so it isn't monotonic, but why the usual method tells us else? We can use it if $a_n\ne0$ for every $n$ which holds here.

答:
Note that the sequence is $\{ a_n\}_{n \in \mathbb{N}} := \{(-3)^n\}_{n \in \mathbb{N}}$. Now, as mentioned earlier, $$ \frac{a_{n+1}}{a_n} < 1 \implies a_{n+1}<a_n \quad \text{For any }n \in \mathbb{N} $$
holds if each $a_i >0$, for any $i \in \mathbb{N}$. But one thing to add here is the behavior of certain sub-sequences of the sequence $\{a_n\}$.

Note that the even subsequence is $\{ a_{2k}\}_{k \in \mathbb{N}} :=\{(-3)^{2k}\}_{k \in \mathbb{N}}=\{3^{2k}\}_{k \in \mathbb{N}}$, increasing.

and the odd subsequence is $\{ a_{2k+1}\}_{k \in \mathbb{N}} :=\{(-3)^{2k+1}\}_{k \in \mathbb{N}}=\{-3^{2k+1}\}_{k \in \mathbb{N}}$, decreasing.

(This observation of the even and odd subsequences come in handy many times, in many problems)


问:
Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$.
I am working on inequality problems for mathematical olympiads. I have come across this problem:

Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $

I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.

I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.

答:
Let $u=\sqrt2x$, $y=v$, $z=w$. Then, it is equivalent to find the maximum $k^2=u^2+v^2+w^2$ subject to
$$\frac u{10\sqrt2} + \frac v{10} + \frac w{10} =1\tag1$$

Observe that $k$ is the distance from the origin to the plane (1) with the axis-intercepts $10\sqrt2,10,10$. In the first quadrant, (1) is a tetrahedron with its furthest corner at the $x$-intercept $10\sqrt2$. Thus, the maximum is $k^2_{max}=(10\sqrt2)^2 =200$.


问:
Semi-metric space with convergence
Let $X$ be a metrical space. Let $B(X)$ be the set of nonempty, bounded subsets of $X$. Define a semi-metrics on $B(X)$ ($0$ distance can exist) which inducates this convergence:
$A_n\rightarrow A$, if for every $\epsilon>0$ there exist $N$ such that for every $n>N$, $A_n\subset A_\epsilon$ and $A\subseteq (A_n)_\epsilon$,
where $H_c$ means the open sets with radius $c$ neighbourhood.

Let $K(X)$ be the set of nonempty, compact subsets of $X$. Show that the prevously defined semi-metrics is a metrics on $K(X)$ and if $X$ is complete, then the defined metric space is complete.

I believe the first part of this question is rather easy, still I was not able to figure it out. Any help would be much appreciated!

答:
Given a metric space $(X,d)$, as I understood your notation the required convergence is provided by Hausdorff distance $d_H$, whose restriction on $K(X)$ is a metric and this is proved in Proposition 2-2 of [H]. When $(X, d)$ is complete, then $(K(X), d_H)$ is complete, and this is proved in Theorem 3-3 of [H].

References

[H] Jeff Henrikson, Completeness and total boundedness of the Hausdorff metric, MIT Undergraduate Journal of Mathematics, 1999 69–80.

Title: Semi-metric space with convergence


问:
Word Problem about no. of square inches of pizza from GRE Quant
Huey's Hip Pizza sells two sizes of square pizzas: a small pizza that measures $10$ inches on a side and costs $\$10$, and a large pizza that measures $15$ inches on a side and costs $\$20$. If two friends go to Huey's with $\$30$ apiece, how many more square inches of pizza can they buy if they pool their money than if they each purchase pizza alone?

My solution: I have solved that problem but however, my answer does not coincide with official answer. Here's my solution:

If they purchase alone they can buy $3$ small pizzas (in that case total area is $100\cdot 3=300$ square inches) or $1$ small+$1$ large (in that case total area is $100+225=325$ square inches). Hence when they purchase each pizza alone then the total area will be $2\cdot 300=600$ square inches or $2\cdot 325=650$.

If they purchase together they can buy:

(i) $6$ small pizzas then total area will be $6\cdot 100=600$

(ii) $1$ large + $4$ small then total area will be $225+4\cdot 100=625$

(iii) $2$ large + $2$ small then total area will be $2\cdot 225+2\cdot 100=650$

(iv) $3$ large then total area will be $3\cdot 225=675$

Eventually, the maximal difference is $675-600=75$.

Is my answer true?

答:
Your assumption about buying $6$ small pizzas is correct but there is no incentive for that .

For $\$30$, each can get a large and a small $\quad \rightarrow 2(225+100)=2\times 325=650$

Combined, they can get $3$ large pizzas $\rightarrow 3(225)=675$

$$675-650=25$$

Your assumption about buying $6$ small pizzas is correct but there is no incentive for that .


问:
Prove that the lim as $x$ approaches $a$ of $x^n$ equals $a^n$. (hint use induction)
Prove that the lim as $x$ approaches $a$ of $x^n$ equals $a^n$. (hint use induction)

I'll be honest and say that I have no idea where to start with this problem. I am familiar with the epsilon delta definition but I am unclear as to how to involve induction and solve the problem.

答:
As the hint is induction, you should rely on $x^n=x^{n-1}x$, and establish that

$$\lim x^{n}=a^n\impliedby \lim (x^{n-1}x)=a^{n-1}a.$$

Then next step in the reasoning is that

$$a^{n-1}a=\lim x^{n-1}\cdot\lim x.$$

You should be able to fill the gap and handle the base case.


问:
Computing the probability to win in a solitaire
A solitaire was played on normal cards such that there are four suits and 13 numbers for each suit. Rules are the following:

You put four cards to front of you. If there are at least two same suits, you remove the cards with same suit and put new cards from the deck to front of you such that there are four cards front of you. If you can't fill the board, you may have 1 to 3 cards at the final round. You lose if the cards in the board are all of different suits at the end. What is the probability that you can remove all the cards in this solitaire? Can this be counted exactly or are there any decent approximation?

答:
I was looking for something else and come up to your question.
I did a small macro in Excel and my results are:

Success rate:  131 out of 1000. , so the probability is ca. 1/7 - 1/8.

You can try to build a decision tree to calculate the real probability.  This should be not very complex as we talk about only 4 colours (cards values don't care).

You can try it in some calculation sheet.


问:
Solving multi DOF damped system using block matrix and matrix exponential
I'm trying to solve a multi DOF damped system for $U$. $[M]$ and $[K]$ are symmetric nxn matrices, but $[C]$ is not symmetric. Can I use the following approach?

Put in block matrix form

Solve the 1st order differential using exponential matrices

Is it true that because $[C]$ is not symmetrical then it can't have distinct eigenvalues/eigenvectors, making it not possible to diagonalize? But even if that's true, $e^{[A]t}$ can still be found without diagonalizing, right?

答:
Your approach seems fine. By differentiating the candidate $X(t)$ we recover the correct differential equation.

A non-symmetric matrix can have distinct eigenvectors/ eigenvalues. Example
$$M= \left(\begin{matrix} 1 & 1 \\ 0 & 2\end{matrix}\right)$$
Is not symmetric but has the following eigensystem
$$\lambda_1=1, u_1=\left(\begin{matrix} 1 \\ 0 \end{matrix}\right)$$
$$\lambda_2=2, u_2=\left(\begin{matrix} 1 \\ 1 \end{matrix}\right)$$

There are many ways to compute the matrix exponential. The one you suggest is perfectly fine. The most direct is by using the power series of the exponential$^\dagger$

$$e^{At}=\sum_{n=0}^\infty \frac{A^n t^n}{n!}$$

Which is useful if your matrix has some nice property of powers, eg if $A^2=1$. A different useful method is via the Cayley-Hamilton theorem, for which you'll only need the eigenvalues. There's a worked example using this on the wiki page.

$\dagger$ Also useful because this definition of the matrix exponential is always well defined (the sum converges).

Answer to question in comment: "why is the damping approximated?"
I can't comment on the engineering textbooks, but I assume the approximation is done in order to get a (set of) linear differential equations, which can then be cast into matrix form. Things are not so simple if your damping term is eg $ u\dot{u}$, the resulting equation is nonlinear and all the nice things done above won't work: the solution can't simply be cast into matrix form. The approximation could also be done to render your matrix $A$ time independent. If $A$ is time dependent (and the differential equations are linear), the method you described won't directly work, but the solution may still be cast in matrix form (see eg. A. Prosperetti: Advanced mathematics, ch18)


问:
Prove that for $f(x) = \frac{x^\textrm{T}Ax}{x^\textrm{T}x}$ and any $v$, $v^\textrm{T}Hv=0$
$f(x) = \frac{x^\textrm{T}Ax} {x^\textrm{T}x}$, where $A$ is a symmetric $n \times n$ matrix and $x \neq 0$. I need to prove that if $H=H(f)(v)$, where $H(f)(v)$ is the Hessian of function $f$ evaluated for vector $v$, $v^\textrm{T}Hv=0$ for any $v$.

I have already proved that:

(a) $\frac{df}{dx}=0$ iff $x$ is an eigenvector of $A$;

(b) $f(cx) = f(x)$ for $c \in R$;

(c) $\lambda_\min \leq f \leq \lambda_\max$ where $\lambda_\max$ and $\lambda_\min$ are the maximal and minimal eigenvalues of $A$.

The hint to the question states the following: “ Hint: do not attempt to calculate the Hessian directly: the algebra is very messy. Use
the intuition gave by (b), that a certain function is constant.”

答:
I don't know how to use the hint.  Perhaps you have some expression involving the Hessian that can be used. Here is a brute-force approach that seems to work for any square matrix $A$ (regardless of symmetry): We have
$$ f(x) = \frac{x^{\top}Ax}{x^{\top}x} $$
Define $N(x)$ and $D(x)$ as numerator and denominator functions:
$$ N(x) = x^{\top}Ax, \quad D(x) = x^{\top}x$$
Then
$$ D(x) f(x) = x^{\top}Ax \quad  $$
Taking the gradient of both sides, expressing the gradient as a row vector, gives
$$ f(x) \nabla D(x)  + D(x) \nabla f(x)  = 2x^{\top}A \quad (*)$$
Now from derivative formulas we know $\nabla D(x) = 2x^{\top}$.
Substituting this into (*) gives
$$ 2f(x) x^{\top}  + D(x) \nabla f(x)  = 2x^{\top} A $$
Taking the derivative again gives
$$ 2(\nabla f(x))^{\top}x^{\top} + 2f(x) I + (\nabla D(x))^{\top} \nabla f(x) + D(x) H(x) = 2A$$
where $I$ is the identity matrix.
Substituting $\nabla D(x)=2x^{\top}$ gives
$$ 2(\nabla f(x))^{\top} x^{\top} + 2f(x) I + 2x \nabla f(x) + D(x) H(x) = 2A$$
Substituting $D(x)=x^{\top} x$, assuming $x \neq 0$, and rearranging terms gives
$$ \boxed{H(x) = 2\left(\frac{A  - f(x) I - x \nabla f(x) -  (\nabla f(x))^{\top}x^{\top}}{x^{\top} x}\right)}$$
Thus
\begin{align}
x^{\top} H(x)x &= 2\left(\frac{x^{\top}Ax  - f(x) x^{\top}x -x^{\top}x \nabla f(x)x -  x^{\top}(\nabla f(x))^{\top}x^{\top}x }{x^{\top} x}\right)\\
& = -2 \nabla f(x) x - 2 x^{\top} (\nabla f(x))^{\top}\\
&= -4 \nabla f(x) x
\end{align}

It remains to show $\nabla f(x) x=0$ for all $x \neq 0$.

$$ f(x) = \frac{N(x)}{D(x)}  $$
$$ \nabla f(x) = \frac{D(x)\nabla N(x) - N(x)\nabla D(x)}{D(x)^2} = \frac{D(x)2x^{\top}A - N(x) 2x^{\top}}{||x||^4}$$
$$ \nabla f(x) x = \frac{D(x)2x^{\top}Ax - N(x) 2x^{\top}x}{||x||^4}=0$$


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答: You are quite right!  It is more or less an arbitrary convention that $i$ is represented by the point $(0,1)$ in the plane, and then necessarily $-i$ has to be represented by $(0,-1)$.  It could have been the other way around!  Regardless of the convention you choose, the PAIR $i, -i$ will always be the same as $(0,\pm 1)$.</p>

Title: What is $-i$ exactly?


问:
iterative logarithm and one asymptotic example
we know about iterative logarithm like as Here.

there are some typo in my note, so I update the question into:
which of them is correct from asymptotic point of view? there is no need to proof, an intuitive idea is enough.

$n\log^*_n = \log^*(n^n)$

$\log^*n = \log^*(n^n)$

Is there a big typo in my notes or really this is true?

答:
$$\log^*(n^n)=\log^*(n\log(n))+1=\log^*(\log(n)+\log(\log(n)))+2\\\sim \log^*(\log(n))+2=\log^*(n)+1\sim\log^*(n).$$

On the other hand,

$$\log^*(n^n)\sim n\log^*(n)$$ cannot be true, as the iterated logarithm is strongly sublinear.


问:
Field extension of $\mathbb{Q}$ is the splitting field of two different polynomials?
On the topic of splitting fields at Wolfram Mathworld, one of the examples makes sense but the other doesn't. From https://mathworld.wolfram.com/SplittingField.html:

For example, the extension field Q(sqrt(3)i) is the splitting field
for x^2+3 since it is the smallest field containing its roots,
sqrt(3)i and -sqrt(3)i. Note that it is also the splitting field for
x^3+1.

The first one makes sense, but I can't figure out how $\mathbb{Q}(\sqrt 3i)$ is the splitting field of $x^3+1$. Any ideas are appreciated.

答: Hint: Write out the three roots of $x^3+1$


答:
Write

\begin{align*}
\omega &= \exp\big(\frac{2 \pi i}{6}\big) \\
& = \frac{1}{2} + \frac{\sqrt{3}}{2}i.
\end{align*}
Then $\omega \in \mathbb{Q}(\sqrt{3}i)$ and $x^3 + 1 = (x+1)(x-\omega)(x+\omega)$ splits in this field.

Title: Field extension of $\mathbb{Q}$ is the splitting field of two different polynomials?

Body:
On the topic of splitting fields at Wolfram Mathworld, one of the examples makes sense but the other doesn't. From https://mathworld.wolfram.com/SplittingField.html:

For example, the extension field Q(sqrt(3)i) is the splitting field
for x^2+3 since it is the smallest field containing its roots,
sqrt(3)i and -sqrt(3)i. Note that it is also the splitting field for
x^3+1.

The first one makes sense, but I can't figure out how $\mathbb{Q}(\sqrt 3i)$ is the splitting field of $x^3+1$. Any ideas are appreciated.


问:
For $z+\frac{1}{z}=2\cos\alpha$ calculate $z^n+\frac{1}{z^n}$ If $n$ is a positive integer
For $z+\frac{1}{z}=2\cos\alpha$ calculate   $z^n+\frac{1}{z^n}$ If $n$ is a positive integer

here's my first attempt

$z^2-2\cos\alpha+1=0$ so we have $z=\cos\alpha\pm i\sin\alpha= e^{\pm i\alpha}$

but I thought there is a simple approach

What other approach would you suggest?

答:
$$\small z+\frac{1}{z}=2\cos{\alpha}\iff\left(z-\cos{\alpha}\right)^{2}+\sin^{2}{\alpha}=0\iff z=\cos{\alpha}\pm\mathrm{i}\sin{\alpha}\iff z=\mathrm{e}^{\pm\mathrm{i}\alpha}\iff z^{n}=\mathrm{e}^{\pm n\mathrm{i}\alpha} $$

Thus : $$ z^{n}+\frac{1}{z^{n}}=\mathrm{e}^{\pm n\mathrm{i}\alpha}+\mathrm{e}^{\mp n\mathrm{i}\alpha}=2\cos{\left(n\alpha\right)} $$


问:
About Ordinary differential equation of first order and first degree.
Q. Show that the differential equation of the form

$[y+xf(x^2+y^2)]dx+[yf(x^2+y^2)-x]dy=0$

is not exact. Also show that $1/(x^2+y^2)$ is an I.F. of DEs of the given form.

Comparing the given DE with $P(x,y)dx+Q(x,y)dy=0,$ Pfaffian differential equation in two variables, it's obvious that

$P(x,y)=y+xf(x^2+y^2)$ and $Q(x,y)=yf(x^2+y^2)-x.$

I know that the necessary and sufficient condition of Pfaffian differential equation to be exact is:

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}.\tag*{}$

So if $\frac{\partial P}{\partial y}\ne\frac{\partial Q}{\partial x}$ then Pfaffian differential equation is not exact.

In the present question,

$\frac{\partial P}{\partial y}=1+x\frac{\partial f}{\partial y}$ and $\frac{\partial Q}{\partial x}=y\frac{\partial f}{\partial x}-1.$

Here how I can ensure that whether $\frac{\partial P}{\partial y}$ and $\frac{\partial Q}{\partial x}$ are equal or not, since the form of $f(x^2+y^2)$ is not known.

I'm not quite sure whether it is a correct approach or not or how to approach further... Please suggest...

答:
Keep in mind that $P$ and $Q$ are functions of two variables, but $f$ is a function of a single variable. So

$$ \frac{\partial P}{\partial y}=1+2xyf^\prime(x^2+y^2) $$

and

$$ \frac{\partial Q}{\partial x}=2xyf^\prime(x^2+y^2)-1. $$

Equality would imply that $1=-1$.


问:
Some questions regarding the $L^2$ space
Let $(\Xi,\boldsymbol{\Xi},\mu)$ be a measure space, and let $(\mathbb{R},\mathcal{B}_\mathbb{R})$ be the Borel space associated to $\mathbb{R}$.
Also, let
$$\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})=\bigg\{f:(\Xi,\boldsymbol{\Xi})\to(\mathbb{R},\mathcal{B}_\mathbb{R})\,\,\Big|\,\int f^2\,\mathrm{d}\mu<\infty\bigg\}$$
be the set of all (real-valued) square-integrable (measurable) functions with respect to (the measure) $\mu$.

Then can we say the following?

When the Lebesgue space $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ is viewed as a set, it is defined as the quotient set:
$$L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})_\text{set}:=\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\,\,/\sim,$$
where $\sim$ is an equivalence relation on $\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ such that:
$$f\sim g\quad:\Leftrightarrow\quad f-g=0\,\text{ $\mu$-almost everywhere.}$$
Is this the definition of $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ as a set?

If so, what does '$\mu$-almost everywhere' actually mean in the above definition?
Some authors seem to imply that (but of course I may be wrong with this interpretation):
$$f(\xi)\neq g(\xi)\quad\Rightarrow\quad \mu(\xi)=0$$
when they say that if $f=g$ then $f$ and $g$ differ only on a set of measure zero. Here $\xi\in\Xi$.
However, for this to work, we require that $\{\xi\}$ is an element of the $\sigma$-algebra $\boldsymbol{\Xi}$ to make sure that $\{\xi\}\mapsto\mu(\xi)$ is well-defined.
This therefore imposes a hidden requirement on the definition of $\boldsymbol{\Xi}$ (which I don't like).
Or is this the correct interpretation?
$$f-g=0\,\text{ $\mu$-almost everywhere}\quad\Leftrightarrow\quad\int (f-g)^2\,\mathrm{d}\mu=0.$$

The reason why we want to define such an equivalence relation $\sim$ on $\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ is that if we equip $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ with the bilinear map:
$$\langle\,\cdot\,,\cdot\,\rangle:L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\times L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\to\mathbb{R}\quad:\Leftrightarrow\quad\langle f,g\rangle=\int fg\,\mathrm{d}\mu,$$
then $\langle\,\cdot\,,\cdot\,\rangle$ qualifies as an inner product on $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ because it satisfies the positive-definite condition of an inner product.
Is this the correct motivation?

$L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ is by definition a vector space over $\mathbb{R}$, and no further structure is assumed on $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})_\text{set}$.

If nothing is said, all (equivalence classes of) measurable functions in $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ are always assumed to be maps from $(\Xi,\boldsymbol{\Xi})$ into the Borel measurable space $(\mathbb{R},\mathcal{B}_\mathbb{R})$, aren't they?

(Disclaimer: I am not a mathematician but an engineer studying functional analysis; thus, the notation used here may be non-standard or simply incorrect. If that is the case, please let me know. I would appreciate it very much.)

Thank you,
Frederick.

答:
I am assuming that by ${\cal L}^2$ you mean the square integrable functions and by $L^2$ you mean the equivalence classes. Both of these are sets and you have $L^2 = {\cal L}^2 / \sim$. (Both are vector spaces with the appropriate definitions of $+,\cdot$.)

$f=g $ $[\mu]$ ae. means there is a set of measure zero (null set) $N$ such that $f(x)=g(x)$ for all $x \notin N$.

The reason for the equivalence classes is so that $f \mapsto \sqrt{\int |f|^2d \mu}$ is a norm
rather than a seminorm.

$L^2$ is a set, I am not sure how you distinguish the two.

No. An element of $L^2$ is the equivalence class of (square integrable) measureable functions $\Xi \to \mathbb{R}$.

In many situations one can be sloppy about the distinction, just as we rarely think of real numbers as equivalence classes. Sometimes one needs to be careful, for example, to define the $\sup$ norm (for $L^\infty$) of support of an equivalence class.


问:
How to prove that the language of Non-palindromes is not regular (via pumping lemma)
I am trying to find that $L ={\{w\text{ | } w ∈ {\{a, b\}} * \text{is not a palindrome}\}}$

This is related to this previous question, though in this case I want to explicitly prove it via pumping lemma, since I am unsure my solution is correct. This is what I have:

$\text{for any } N, \text{ let } w \in L, |w|\ge N$

$w=a^Nb^Na^{N+1}$

$\text{For any partition } w =xyz , |y|\ge 1, |xy| \le N$

$x=a^r, y=a^s, z=a^{N-r-s}b^Na^{N+1},\text{where }s\ge 1$

$\exists i \text{ where } xy^iz\notin L$

$i=2\text{ ---> }xy^2z=a^{N+s}b^Na^{N+1} \text{ ,where we know that } s\ge 1$

$\text{if }s=1 \text{ then }xy^2z\notin L \text{ and thus L is not regular}$

I am really unsure about the last part. Does the pumping lemma hold if the "pumped word" doesn't belong to L for some specific values of $s$?

答:
There are several problems here. First, you can’t use just any $N$: you need to specify that $N$ is at least as big as the pumping length. Next, the pumping lemma says that there is at least one partition $w=xyz$ such that $|y|\ge 1$, $|xy|\le N$, and $xy^iz\in L$ for each $i\ge 0$; it does not say that this is the case for every partition $w=xyz$ with $|y|\ge 1$ and $|xy|\le N$. And that is why it is indeed illegitimate to pick a specific value of $s$: the pumping lemma just says that there is at least one pair of $r$ and $s$ such that $r+s\le N$, $s\ge 1$, and

$$a^{r+is}a^{N-r-s}b^Na^{N+1}=a^{N+(i-1)s}b^Na^{N+1}\tag{1}$$

is in $L$ for each $i\ge 0$. There is no guarantee that $s=1$. And since $s=1$ is the only value of $s$ for which we can find an $i\ge 0$ making $(1)$ a palindrome, your choice of $w$ simply can’t be made to work.

Because we have no control over $|y|$ beyond the fact that $1\le|y|\le N$, I see no way to use the pumping lemma to show directly that $L$ is not regular. You can use it to show that the language of palindromes over $\{a,b\}^*$, which is the complement of $L$, is not regular and then use the fact that a language is regular if and only if it’s complement is regular to conclude that $L$ cannot be regular. Alternatively, you can use the Myhill-Nerode theorem, as in the answer to this question.


问:
Shared probability function of dependent Poisson variable and Binomial variable
The number of people that enter a shop is a Poisson variable with parameter $\lambda=50$. The probability that a customer that entered the shop will buy something is $p=0.16$.

I need to calculate the probability that:

exactly 20 customers, out of all the customers that entered the shop in 2 hours, will buy something

exactly 5 customers buy something if it is known that in 1 hour 40 customers entered the shop

let $X$ be the number of customers who buy something, and let $Y$ be the number of customers that entered the shop. The shared probability function $p(x, y)$ when $x=20$ is $$p(20, y)=e^{-100}\frac{100^y}{y!}\binom{y}{20}(0.16)^{20}{0.84}^{y-20}$$
hence the required probability is $$\sum_{y=20}^{\infty}e^{-100}\frac{100^y}{y!}\binom{y}{20}(0.16)^{20}{0.84}^{y-20}=\sum_{y=20}^{\infty}e^{-100}\frac{100^y}{y!}\frac{y!}{20!(y-20)!}(\frac{0.16}{0.84})^{20}{0.84}^y=\frac{e^{-100}}{20!}(\frac{0.16}{0.84})^{20}\sum_{y=20}^{\infty}\frac{100^y \cdot 0.84^y}{(y-20)!}=\frac{e^{-100}}{20!}(\frac{0.16}{0.84})^{20}\sum_{y=0}^{\infty}\frac{84^{y+20}}{y!}=\frac{e^{-100}}{20!}(16)^{20}\sum_{y=0}^{\infty}\frac{84^y}{y!}=\frac{e^{-100}}{20!}(16)^{20}e^{84}=\frac{e^{-16}}{20!}(16)^{20}\approx 0.0559$$

2.

$$P\{X=5|Y=40\}=\frac{P\{X=5, Y=40\} }{P\{Y=40\} } = \frac{e^{-50}\frac{50^{40}}{40!}\binom{40}{5}(0.16)^{5}{0.84}^{35}}{e^{-50}\frac{50^{40}}{40!}}=\binom{40}{5}(0.16)^{5}{0.84}^{35}\approx 0.1544$$

But I've a problem settling these results with what I understand, it appears, to me, in both answers that $X$ and $Y$ are independent, while I know that they aren't.

答:
Let $Y$ be the number of people that entered the shop, and let $X$ be the number of people who further bought something.

You know:

$Y \sim \text{Poisson}(\lambda)$.

$X = \sum_{i=1}^Y Z_i$, where $(Z_i)$ are independent of $Y$ and iid with distribution $P(Z_i = 1) = 1 - P(Z_i = 0) = p$.

In other words, given $Y = y$, X follows a $\text{Bin}(y,p)$-distribution.

The simultaneous probability function $p(x,y)$ may be written as
$$
p(x,y) = P(X = x \, | \, Y = y) \cdot P(Y = y)
$$
for $(x,y) \in \{(n,m) \in \mathbb{N}_0 : n\leq m\}$. (It is zero otherwise.) By inserting the relevant marginal and conditional distribution, one arrives at your expression.

Does this answer your question?


问:
Prove that $ (A \cup B)'\subset A' \cup B'$
Prove that $ (A \cup B)'\subset A' \cup B'$

My attempt:

Suppose $p\in (A\cup B)'$, If $p\notin A'$.claim that  $p\in B'$. since $p\notin A'$ so there exist  $\epsilon_0>0$ such that $N_{\epsilon_0}(p)\cap A \subset \{p\}$.

Now for any $\epsilon >0$ ,let $r= \min \{\epsilon,\epsilon_0\}$. Since $p \in (A \cup B)'$, $N_r(p)\cap A\subset \{p\}$. From the assumption that $p\in(A\cup B)'$, we deduce that $N_r(p)\cap(A\cup B)\ne\emptyset$, so $N_r(p)\cap B\ne\emptyset$.

It follow that $p$ is a limit point of $B$, so $p \in B'= (A\cup B)'$. This implies that $(A \cup B)'= B'$.

Finally $(A \cup B)'\subset  A' \cup B'$ which completes the proof.

Note : $N_r(p)$ denote the neighborhood of $p$ consisting of all $q$ such that $d(p,q)  <r$ and  $A'$ denote  the set of all derived points of $A$

Is this proof correct or not?

答:
To clarify the picture, let us recall the definition of "derived
point" or "accumulation point". Let $(X,\tau)$ be a topological
space. Let $A\subseteq X$. We say that $p\in X$ is an accumulation
of $A$ if for each neighborhood $U$ of $p$, $(U\setminus\{p\})\cap A\neq\emptyset.$
The set of all accumulation points of $A$ is denoted by $A'$.

Let $A,B\subseteq X$, we go to prove that $(A\cup B)'\subseteq A'\cup B'.$
Prove by contradiction. Suppose the contrary that there exists $p\in(A\cup B)'\setminus(A'\cup B').$
Hence, $p\notin A'$ and $p\notin B'$. Choose an open neighborhood
$U$ of $p$ such that $U_{p}\cap A=\emptyset$, where $U_{p}:=U\setminus\{p\}$.
Choose open neighborhood $V$ of $p$ such that $V_{p}\cap B=\emptyset$.
Let $W=U\cap V$, which is an open neighborhood of $p$. Observe that
$W_{p}\cap A\subseteq U_{p}\cap A=\emptyset$ and $W_{p}\cap B\subseteq V_{p}\cap B=\emptyset$.
Therefore, $W_{p}\cap(A\cup B)=(W_{p}\cap A)\cup(W_{p}\cap B)=\emptyset$,
which contradicts to the fact that $p\in(A\cup B)'.$


问:
Assume $T$ is a linear operator over a vector space $V$ with a finite dimension and assume a linear operator $U$ on $V$ does exist such that $TU=I$
Assume $T$ is a linear operator over a vector space $V$ with a finite dimension and assume a linear operator $U$ on $V$ does exist such that $TU=I$,where $I$ is the identity linear transformation.

Show that $T$ is  invertible and that $U=T^{-1}$.

From the assumption we have that $TU=I$ and it follows that $U$ is surjection and $T$ is injection,moreover we have the following lemma:

If $V$ and $W$ are finite vectors spaces over the field $\mathbb F$ such that $\text{dim}(V)=\text{dim}(W)$ and $T:V \to W$ is a linear transformation then the following statements are equivalent:

$T$ is invertible

$T$ is nonsingular,that is $T( \alpha)=0_w \implies  \alpha=0_v$

$T$ is surjection

So now from the lemma we know that $T$ has an inverse $T^{-1}$ such that $TT^{-1}=I=TU$,the linear transformation $T$ is well-defined and so $T^{-1}TT^{-1}=T^{-1}TU \implies IT^{-1}=IU \implies T^{-1}=U$.

I used the fact that $T$ is well-defied,but is it really well-defined? since the mapping has not been given explicitly then I think it's kinda meaningless to talk about the well-definedness,however without that I'm not able to finish the proof.

答:
From the assumption we have that $TU=I$ and it follows that $U$ is surjection and $V$ is injection,...

That's not true, $U$ is an injection since $$(\forall v_1,v_2 \in V) : \ Uv_1=Uv_2 \ \Rightarrow \  v_1 = T(Uv_1) = T(Uv_2) = v_2$$ and $T$ (not $V$) is a surjection since $(\forall v \in V) \ v = T(Uv)$. Anyway, you can also apply the lemma and continue with the proof.

So now from the lemma we now that $T$ has an inverse $T^{−1}$ such that $TT^{−1}=I=TU$, the linear transformation $T$ is well-defined and so...

I don't see the necessity of saying that $T$ is well-defined. What is a well-defined function for you? In this problem we don't need an explicit formula/rule for $T$, it is just an abstract linear map that we are assuming that has a right inverse. So, if, in the future, you are working in a concrete problem and appears a linear operator defined in a finite-dimensional vector space that has a right inverse, this exercise/result shows you that linear operator is necessarily invertible.


问:
Verify proof that limit ordinal iff $\alpha = \cup_{\beta<\alpha} \beta$.
I'm filling in the gaps of the explanations of a logic textbook which has a very terse treatment of ordinals, so I was hoping someone could take a look at my proof of:

Proposition: Let $\alpha$ be an ordinal. Then $\alpha \neq \beta+1$ for all $\beta$ and $\alpha \neq \varnothing$ if and only if:

$$\alpha = \bigcup_{\beta < \alpha} \beta$$

Here is my proof:

Let $\alpha$ be an ordinal such that $\alpha \neq \beta + 1$ for all ordinals $\beta$ and $\alpha \neq \varnothing$. Proven in the textbook is that either $\alpha \in \beta$ or $\alpha = \beta$ for all ordinals, so it must be the case that for all $\beta + 1 < \alpha$, we have $\beta + 1 \in \alpha$. This is the same as $\beta \cup \{\beta\} \in \alpha$.

It follows that:

$$\bigcup_{\beta < \alpha} \beta \subseteq\bigcup_{\beta < \alpha} (\beta \cup \{\beta\}) \subseteq \alpha$$

since set membership holds for every $\beta < \alpha$. Now let $\gamma \in \alpha$, hence $\gamma < \alpha$; as $\gamma$ is a well-ordered set, it must be isomorphic to an initial segment (all elements less than) of $\{\beta, \{\beta\}\}$, proved in the textbook, hence $\gamma = \beta$ for a unique $\beta$ and $\alpha \subseteq \bigcup_{\beta < \alpha} \beta$.

For the other direction:

Let $\alpha$ be an ordinal such that $\alpha = \bigcup_{\beta < \alpha} \beta$. Assume for contradiction that $\alpha = \beta + 1$ for some $\beta < \alpha$. This must be the $\beta$ covered by $\alpha$, for if there was some $\beta < \beta + 1 < \alpha$, we would have $\alpha < \alpha$. Then:

$$\beta \cup \{\beta\} = \bigcup_{\beta < \alpha} \beta$$

Which implies that:

$$\{\beta\} = \bigcup_{\gamma < \beta} \gamma$$

However, the right-hand side, by the definition of an ordinal, is $\beta$ itself, so we have $\beta = \{\beta\}$, a contradiction.
$\square$

In particular, I am somewhat confused about how to show that if $\alpha$ may be written as a union, then it is not $\varnothing$. Can't we say that $\varnothing = \cup_{\beta < \varnothing} \beta$ trivially?

答:
You're exactly right about $\varnothing$ -- it does satisfy $\varnothing = \bigcup_{\beta < \varnothing} \beta$, and $\varnothing$ is not a limit ordinal! I think the correct claim is:

Let $\alpha$ be an ordinal. Then $\alpha$ is not a successor ordinal (i.e. for all $\beta$, $\alpha \neq \beta + 1$) if and only if $\alpha = \bigcup_{\beta < \alpha} \beta$.

or

Let $\alpha$ be an ordinal. Then $\alpha$ a limit ordinal if and only if $\alpha = \bigcup_{\beta < \alpha} \beta$ and $\alpha \neq 0$.

These are equivalent, so I'll prove the first version. Before I do, some comments on your proof:

[Forwards Direction]

"Proven in the textbook is that either $\alpha \in \beta$ or $\alpha = \beta$" is a bad way to say this. It is not true that for all ordinals $\alpha$ and $\beta$, $\alpha \in \beta$ or $\alpha = \beta$. For example, let $\alpha = 1$ and $\beta = 0$. The correct statement is that $\alpha \in \beta$ OR $\alpha = \beta$ OR $\beta \in \alpha$. In other words, we have $\alpha < \beta$, or $\alpha = \beta$, or $\beta < \alpha$: the ordinals are totally ordered.

Did you mean to say that $\beta < \alpha \implies \beta + 1 \in \alpha$? What you said is also true, but less useful.

The fact that $\bigcup_{\beta < \alpha} \beta \subseteq \alpha$ is simpler than you make it out to be: this is just a union of subsets of $\alpha$ (ordinals are transitive sets, so $\beta < \alpha \implies \beta \in \alpha \implies \beta \subseteq \alpha$).

The rest of this part of the proof doesn't make sense to me. Notably, you never used the fact that $\alpha$ is not a successor ordinal! Also, it's unclear what $\beta$ is supposed to be. Instead, to show $\alpha \subseteq \bigcup_{\beta < \alpha} \beta$, we take an arbitrary $\gamma \in \alpha$ and note that $\gamma \in \gamma + 1$. Since $\alpha \neq \gamma + 1$, we have $\alpha < \gamma + 1$ or $\gamma + 1 < \alpha$. If $\alpha \in \gamma + 1 = \gamma \cup \{\gamma\}$, then $\alpha \in \gamma$ (but then $\gamma \in \alpha \in \gamma \implies \gamma \in \gamma$, contradiction!) or $\alpha = \gamma$ (but then $\gamma \in \gamma$, contradiction!). Thus, $\gamma + 1 < \alpha$, so $\gamma \in \gamma + 1 \subseteq \bigcup_{\beta < \alpha} \beta$.

[Reverse Direction]

What does "the $\beta$ covered by $\alpha$" mean? This is unclear, and (I think as a result) I don't understand the rest of that sentence.

The line $\beta \cup \{\beta\} = \bigcup_{\beta < \alpha} \beta$ is confusing because the $\beta$'s on the left- and right-hand sides mean different things. It would be better to write $\beta \cup \{\beta\} = \bigcup_{\gamma < \beta + 1} \gamma$, or something like that.

It's unclear how you're getting $\{\beta\} = \bigcup_{\gamma < \beta} \gamma$. I think the argument you're looking for should go something like "$\{\beta\} = (\beta \cup \{\beta\}) \setminus \beta = (\bigcup_{\gamma < \beta + 1} \gamma) \setminus \beta =$...", but you need to fill this in -- I don't see how to derive this at the moment.

The right-hand side is not equal to $\beta$ by definition -- I think you're thinking of the fact that $\beta = \{\gamma : \gamma < \beta\}$. The correct thing to say would be that the right-hand is equal to $\bigcup \beta$. The whole point of this exercise is that some ordinals are equal to their unions and some are not!

Ok, now here's a proof of the claim:

First, suppose $\alpha$ is a successor ordinal; let $\alpha = \beta + 1 = \beta \cup \{\beta\}$. Note that $\beta \in \alpha$. Now let $\gamma < \alpha$ be arbitrary. Then either $\gamma \in \beta$ or $\gamma = \beta$. Either way, we must have $\beta \notin \gamma$. Thus, $\beta \in \alpha \setminus \bigcup_{\gamma < \alpha} \gamma$, so $\alpha \neq \bigcup_{\gamma < \alpha} \gamma$.

Next, suppose $\alpha \neq \bigcup_{\gamma < \alpha} \gamma$. Clearly $\bigcup_{\gamma < \alpha} \gamma \subseteq \alpha$, so we can pick an element $\beta \in \alpha \setminus \bigcup_{\gamma < \alpha} \gamma$. Now let $\gamma < \alpha$ be arbitrary. Since $\beta \notin \gamma$, we have $\gamma = \beta$ or $\gamma \in \beta$. Either way, $\gamma \in \beta \cup \{\beta\} = \beta + 1$. Since $\gamma$ was arbitrary, $\alpha \subseteq \beta + 1$. Also, $\beta \in \alpha$ implies $\beta + 1 \subseteq \alpha$, so we conclude that $\alpha = \beta + 1$, and thus $\alpha$ is a successor ordinal.


问:
Distribution of a division of two absolutely continuous random variables
Assume $Z,Y $ are independent standard normal random variables.

Find the distribution of $ Z/Y $.

The answer is that $ Z/Y $ is absolutely continuous random variable with density

$$ f_{\frac{Z}{Y}}(x)=\frac{1}{\pi(1+x^2)} $$

Here's what I don't understand:

I tried to write:

\begin{align}
F_{\frac{Z}{Y}}(t) & =\mathbb{P}\left(\frac{Z}{Y}\leq t\right)=\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y<0\right)+\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y>0\right) \\[8pt]
& =\mathbb{P} (Z\geq tY\cap Y<0)+\mathbb{P}\left(Z\leq tY\cap Y>0\right) \\[8pt]
& =\mathbb{P}(Z\geq tY) \mathbb{P}(Y<0)+\mathbb{P}(Z\leq tY) \mathbb{P}(Y>0) \\[8pt]
& =\frac{1}{2}\mathbb{P}(Z\geq tY)+\frac{1}{2}\mathbb{P} (Z\leq tY)
\end{align}

And then to calculate each term. for example :

$$ \mathbb{P}(Z\leq tY)=\mathbb{P} ((Z,Y)\in \{ (z,y):z\leq ty \} ) = \intop_{-\infty}^\infty \intop_{-\infty}^{ty} \frac{1}{2\pi} e^{-\frac{z^2+y^2}{2}} \, dz\, dy $$

But this leads me to a wrong answer, and I'm pretty sure that the problem is somewhere in what I wrote before.

Any help would be appreciated, thanks in advance.

答: Draw a picture of a $(y,z)$-plane and sketch the region consisting of points $(y,z)$ satisfying $z/y \le t$. This should look like two cones (or pizza slices) with points at the origin and extending infinitely away from the origin. Because the distribution of $(Y,Z)$ is rotationally symmetric, the probability that $(Y,Z)$ lies in this region is proportional to the total angle that these cones sweep out. The angle in question will involve arctangent and $t$, and differentiating to get the PDF will lead to the $\frac{1}{1+t^2}$ expression.


问:
Proving basic facts about sets and proper classes
I have the following statements:

Definition of set. $y$ is a set if and only if there is $z$ such that $y\in z$; otherwise $y$ is a proper class

The comprehension axiom. Given a formula $\psi(x)$ where $x$ occurs free in $\psi$ and a class $y$, we have that $y\in\{x:\psi(x)\}$ if and only if $y$ is a set and $\psi(y)$

Definition of the empty set. $\emptyset$ is defined as $\{x:x\neq x\}$

Definition of the universe. $V$ is defined as $\{x:x=x\}$

At this point I managed to prove that:

no set belongs to $\emptyset$

every set belongs to $V$

$\{x:x\notin x\}$ is a proper class

How can I prove that $\emptyset$ is a set and that $V$ is a proper class? How can I prove that $\{x:x\notin x\}$ is equal to $V$?

Do I need additional statements perhaps?

Thank you.

答:
To show the empty set is a set use the power set axiom, as it appears in every power set.

To show the universe is not a set, again use the power axiom and deduce from that $V\in V$ to derive a contradiction.

Edit:
We first prove some lemma, suppose $M$ is a proper class then it is non-empty.

Indeed, assume $M$ is an empty class, take any $x$ then $\{y\in x\mid y\in M\}$ is a subset of $x$, i.e. an element of the power set of $x$, which is the set of all subsets of $x$.

By extensionality we can now prove that this is the empty set.

The others two questions are of a similar state of mind, using the axiom of foundation if you prefer to avoid the power axiom.


答:
After defining a new symbol $i$ such that its square is the negative number $-1$, you can do algebra with it. So, you have an additive operation on the set of symbols like $a+bi$, with $a,b\in \mathbb{R}$. But, if you remember, any element must have an inverse additive and you denote it by $-i$.

Edit:

ps: I suggest you to read about the construction of complex numbers field. How is it defined?

ps1: when you talk about vectors $(a,b)$ (nothing more than ordered pairs) you are identifying $\mathbb{C}$ with $\mathbb{R}^2$.

Title: What is $-i$ exactly?

Body:
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?


问:
How to get the floor function as a Mellin inverse of the Hadamard product of the Riemann zeta function?
The floor function is given - by Perron's formula - as a Mellin inverse of the zeta function. namely :
$$\left \lfloor x \right \rfloor=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(s)\frac{x^{s}}{s}ds\;\;\;(c>1)$$
This is easily proven using the Dirichlet series rep. of the zeta function : $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$.
i was wondering if one can obtain the same result using the  Hadamard product rep. :
$$\zeta(s)=\pi^{s/2}\frac{\Pi_{\rho}\left(1-\frac{s}{\rho}\right)}{2(s-1)\Gamma\left(1+\frac{s}{2}\right)}$$ </p>

答:
Well, you can, somewhat indirectly. Riemann originally developed the intermediary step function $J(x)$, with steps of $1/j$ at each prime power $p^j$, to establish the prime number theorem. The function $J(x)$ is directly derived from $\ln \zeta(s)$, where he decomposed the Hadamard product into a sum under the natural logarithm (see "Riemann's Zeta Function," H.M. Edwards; I have the Dover edition, 2001).

In that context, the whole number staircase function (sometimes called the integer floor function, or natural number counting function) may be expressed as,
$$
N(w) = \int_{r=1^-}^{w} \exp^*(dJ(e^v))[\ln r] dr,
$$
where $\ln r$ is the independent variable of the convolution exponential, and the argument of the c.e. is just Riemann's $J(x)$:
$$
dJ(e^u) = \left( \frac{1}{u} - \sum_{\rho} \frac{e^{(\rho-1)u}}{u} - \frac{1}{e^{u}(e^{2u}-1)u} \right),
$$
where the $\rho$ are the non-trivial zeros of the Riemann zeta function.

To show it:
$$
N(w) = \frac{1}{2 \pi i} \int_{x-i\infty}^{x+i\infty} \zeta(z) \frac{w^z}{z} dz \Rightarrow N(e^u) = \frac{1}{2 \pi i}  \int_{x-i\infty}^{x+i\infty}  \zeta(z) \frac{e^{uz}}{z} dz.
$$
We'll drop the $1/(2\pi i)$ factor in some of what follows. Then rewrite,
$$
N(e^u) = \int_{x-i\infty}^{x+i\infty} e^{\ln \zeta(z)} \frac{e^{uz}}{z} dz = \int_{x-i\infty}^{x+i\infty} \sum_{k=0}^{\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz.
$$
If we can integrate termwise (*), we've
$$
= \sum_{k=0}^{\infty} \int_{x-i\infty}^{x+i\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz.
$$
Note, each integral is just Perron's formula applied to the Dirichlet series,
$$
D^k = (\ln \zeta(z))^k = \left( \sum_i p_i^{-z} + (1/2)p_i^{-2z} + (1/3)p_i^{-3z} + ... \right)^k
$$
where $i$ indexes the primes. We can take the derivative of the Perron step function to obtain a distribution of point masses of weight $c/(j_1\cdots j_k)$ at each $p_{i_1}^{j_1} \cdots p_{i_k}^{j_k}$. We can apply the Laplace transform convolution formula (in the sense of distributions):
$$
\int_{0^-}^{u} \int_{x-i\infty}^{x+i\infty} \frac{D^k}{k!} e^{tz} dz dt = \int_{0^-}^{u} (1/k!) (\mathcal{L}^{-1}(D))^{*k} [t] dt,
$$
where $t$ is the post-convolution independent variable. The whole series can then be expressed as a convolution exponential,
$$
N(e^u) = \int_{0^-}^{u} \exp^*(\mathcal{L}^{-1}(D))[t] dt,
$$
where the $u$-integral swap with the exponential series sum is justified by noting $\exp^*(.)$ is just a point measure on $[0,\infty)$, with finitely many delta masses in $[0,u]$.

The argument of the convolution exponential has a common form,
$$
\frac{1}{2\pi i} \int_{x-i\infty}^{x+i\infty} \ln(\zeta(z)) e^{uz} dz = dJ(e^u)e^u,
$$
where $J(x)$ is from Edwards.

The outer $e^u$ term can be passed through the convolution exponential, and we can change variables back, $u=\ln w$, and let $t=\ln r$ for the variable of integration,
$$
N(w) = \int_{t=0^-}^{\ln w} \exp^*(e^v dJ(e^v))[t] dt = \int_{r=1^-}^{w} r \exp^*(dJ(e^v))[\ln r] \frac{dr}{r}.
$$

The convolution exponential seems a bit underserved in the literature. I suppose it's assumed things can be done in the frequency domain. It has some of the expected properties, like for functions $f$ and $g$ under convergence conditions, $\exp^*(f+g) = \exp^*(f)*\exp^*(g)$, and if $0<f<g$, then $\exp^*(f)<\exp^*(g)$.

(*) We need to pass the integral inside the sum in,
$$
N(e^u) = \int_{x-i\infty}^{x+i\infty} \sum_{k=0}^{\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz.
$$
The justification is indifferent to finitely many terms of the sum. From the Dirichlet sum, $D$, $(\ln \zeta(z))^k$ has the lowest term in its Dirichlet series as $2^{-kz}$ ($2$ being the lowest prime). Take $K = \mbox{ceil}(e^u/2)+1$. We'll show the integral sum swap works for
$$
\int_{x-i\infty}^{x+i\infty} \sum_{k=K}^{\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz
$$

Note by Perron's formula, we can integrate the relevant Dirichlet series termwise to give,
$$
\int_{x-i\infty}^{x+i\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz = 0,
$$
for all $k\ge K$. To show the swap, we are then concerned with the rate at which the partial sum of the improper-integrals converge to $0$.

Since $\zeta(z)$ is bounded away from $0$ and $\infty$ along the integration contour ($x>1$), so is $\ln \zeta(z)$. Since on finite vertical contours $[x-ih,x+ih]$ the partial sums of the functions converge uniformly, we can pass this "middle" portion of the contour inside the infinite sum--we need only be concerned with the "tails" of the contours:
$$
\int_{x+ih}^{x+i\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{wz}}{z} dz,
$$
(likewise for $\rightarrow x-i\infty$). These actually behave nicely.

In fact, from Edwards (Section 3.5),
$$
\left| \int_{x+ih}^{x+i\infty} \left( \frac{w}{n} \right)^z \frac{dz}{z} \right| \le c \frac{1}{n^x h}
$$
We apply this termwise to the terms in the $(\ln \zeta(z))^k$ Dirichlet series (again, by standard Perron formula results, this swap is OK to do), treating $e^u$ as $w$ to get
$$
\left| \int_{x+ih}^{x+i\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz \right| \le \frac{c_2^k}{h k!},
$$
where,
$$
\left| \sum_i p_i^{-z} + (1/2)p_i^{-2z} + (1/3)p_i^{-3z} + ... \right| \le \sum_i p_i^{-x} + (1/2)p_i^{-2x} + (1/3)p_i^{-3x} + ...  \equiv c_2
$$
since $x>1$ and everything is absolutely convergent. Since the partial sum of the tails converge $\mathcal{O}(1/h$), this justifies the swap.


问:
Using flux to calculate concentration in a given region
I am told that the diffusion equation for radially symmetric flow is $\frac{∂u}{∂t}=\frac{1}{r}\frac{∂}{∂r}\left(rD\frac{∂u}{∂r}\right)$. This is with flux $J=-D\frac{∂u}{∂r}$.

I can show that the steady state of this equation, with boundary conditions $u(a)=0$ and $u(b)=u_0$, is $u(r)=\frac{u_0}{log\left(\frac{b}{a}\right)}log\left(\frac{r}{a}\right)$. So for this steady state the flux is $J=-D\frac{∂u}{∂r}=-\frac{Du_0}{rlog\left(\frac{b}{a}\right)}$.

I am now asked to find the total amount of material in the region $a\leq r\leq b$ but I am confused how to do this, or what it has to do with the flux I just calculated.

答:
Assuming that $u(r)$ refers to the concentration of the material in question, then all you have to do is integrate your expression for your steady-state $u(r)$ in the volume bounded by $a<r<b$.

The calculation has nothing to do with the flux of material exiting or entering the volume—the flux expression you find only tells you that those fluxes are the fluxes required to hold the concentration of material constant over time within that region.


问:
Define operations of cutting and pasting in topology, formally.
In topology a lot of arguments require to cut a manifold along a non-trivial closed simple curve, which gives other manifolds, and pasting manifolds along their boundaries.

How are those operations defined in a more formal way, for example given a manifold $M$ then $$M=A *_C B $$ for some manifolds $A$ and $B$, is the cutting of $M$ along the curve $C$ which gives $A$ and $B$, what is the operation $*_C$?

Similarly for pasting given two manifolds $A'$,$B'$ with boundaries $C_1,C_2$ respectively, then $$A' *_{C_1 \sim C_2}'B'=M'$$ is the identification of their boundaries, what is the operation $*_{C_1 \sim C_2}$ ?

答:
First of all, you should decide what you mean by a "manifold:" There are topological manifolds, PL (piecewise-linear) manifolds, differentiable manifolds. All these have slightly different notions of "cut-and-paste" operations. I will restrict to "topological manifolds."

If $C$ is a Jordan curve in a surface $M$, one of two things will happen:

(a) $C$ is 2-sided.

(b) $C$ is 1-sided.

The former means that $C$ has a closed "product neighborhood" in $S$, which is a compact subset $N(C)$ in $M$ homeomorphic to the annulus $S^1\times [-1,1]$ such that the homeomorphism sends $C$ to the circle $S^1\times \{0\}$.

In the latter case, $N(C)$ is a Moebius band and $C$ corresponds to the "core curve" of the band.

The fact that each Jordan curve is either of type (a) or type (b) is a nontrivial result (a form of Schoenflies theorem). A similar result fails (in general) for hypersurfaces $S$ in manifolds $M$ of dimension $\ge 3$, but holds under an extra condition, called tameness of $S$. (Personally, I prefer to think of tameness as a part of the definition of a topological submanifold, but I am a minority here.) Under this condition, $N(C)$ is again either the product $S\times [-1,1]$ or is the total space of a nontrivial interval bundle over $S$, which is a higher-dimensional analogue of the Moebius band (there are, typically, infinitely many interval bundles).

Now, the operation of "cutting $M$ along $C$" means removing from $M$ the interior of $N(C)$. The result is a surface with one or two boundary curves. It is sometimes denoted $M\backslash \! \backslash C$.

Now, to the converse: Suppose that $M_1, M_2$ are two surfaces with boundary and $C_i\subset \partial M_i$ are boundary components which are closed (Jordan) curves. Pick a homeomorphism $f: C_1\to C_2$: Up to isotopy there are exactly two such maps. If one thinks of $C_2$ as the unit circle in the complex plane, then one obtains the second map $\bar{f}$ from the first one by complex conjugation.

Then one defines the adjunction space
$$
X_f=M_1\cup_f M_2.
$$
Informally, you glue $M_1, M_2$ via $f$. The formal definition is that you first take the disjoint union $X=M_1\sqcup M_2$ and then introduce an equivalence relation on  $X$ by: $x\sim f(x), x\in C_1$. (For the points $x\notin C_1\sqcup C_2$ the equivalence class of $x$ consists only of the point $x$.) Lastly, form the quotient space $X_f:=X/\sim$ with the quotient topology.

It turns out that (in the case of surfaces) the homeomorphism class of $X_f$ is independent of the choice of $f$. A similar construction works if $M=M_1=M_2$ and $C_1, C_2$ are (possibly equal) boundary components of $M$. In the case when $C_1=C_2$ one adds the extra requirement that $f(x)\ne x$ for all $x\in C_1$.

Then, in general, $X_f$ depends on the isotopy class of $f$ and, up to a homeomorphism, there are at most two surfaces obtained this way. The space $X_f$ is always a (topological) manifold, possibly with boundary, and $\partial X_f$ is the image of $\partial X - (C_1\cup C_2)$ under the quotient map.

The pasting construction also works in higher dimensions, but, now, there could be more than two isotopy classes of homeomorphisms and, accordingly, more than two homeomorphism classes of manifolds $X_f$.


问:
Is the following gamma function integral correct?
$$\int_{0}^{\infty}\frac{x^{v-1}}{e^{-\beta\mu}{e^x}-1}=\Gamma(v)\left(e^{\beta\mu}+\frac{e^{2\beta\mu}}{2^v}+\frac{e^{3\beta\mu}}{3^v}+\dotsb\right)$$

This integral is given in my statistical physics assignment and I feel that the $\mu$ in the integrand should have the opposite sign (as my results would make more sense) but I don't know how to prove this and I can't find the formula anywhere. I would love to find a source for this formula.

答:
On the RHS you have:
$$\Gamma(v)\sum_{k=0}^\infty\frac{(e^{-\beta\mu})^k}{k^v}\tag{1}$$
we know that:
$$\Gamma(v)=\int_0^\infty x^{v-1}e^{-x}dx$$
$$\sum_{k=0}^\infty\frac{(e^{-\beta\mu})^k}{k^v}=\Phi(e^{-\beta\mu},v,0)$$
I have changed it to a negative sign because it is the only way that this series will converge for $\beta,\mu>0$ and $\Phi$ represents the Lerch transcendent
. On the same page you will find the identity:
$$\Phi(z,s,a)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{t^{s-1}e^{-at}}{1-ze^{-t}}dt$$
which we can rearrange to give us the general formula:
$$\int_0^\infty\frac{t^{s-1}e^{-at}}{1-ze^{-t}}dt=\Gamma(s)\Phi(z,s,a)$$

Try looking into these functions and seeing if there is any material on this derivation. Note: It is interesting that this summation (with $a=0$) is close in appearance to the Riemann zeta function


问:
Help define this recursive sequence iteratively
(Note you do not need to have seen the original olympiad problem to answer my question)

I was attempting the fifth problem from the 2008 Canadian Mathematical Olympiad, and got all the way to this line in the proof ($r$ is a sequence of numbers, $r_1=1$ and $r_2=4$):

$$ r_{n+1} + 1 = 2(r_{n}+1)+ (r_{n-1}+1)$$

The next line of the provided solution then goes to this:

$$ r_{n} + 1 = {1\over 2\sqrt{2}}(1+\sqrt{2})^{n+1} - {1\over 2\sqrt{2}}(1-\sqrt{2})^{n+1}$$

While I see how this line could be proved via induction, I do not understand how they have got from the first line to the second in the first place; can anyone provide an 'intuitive' way to get to the iterative version of a recursive function, and perhaps how to do so more generally as well? Ideally answers would not use concepts much beyond high school math since after all this is a high school olympiad.

All answers are much appreciated.

答:
Let $u_n = r_n+1$ like the other solution, which is motivated by the fact that no $r_i$ is not without a +1.

Our relation is $$u_{n+1} = 2u_n + u_{n-1}.$$

Our goal is to find an expression that gives us the closed form. In general, we can  create the characteristic polynomial of a relation $d_n =  ad_{n-1} + bd_{n-1}$ as $x^2 -ax - b = 0.$ When the roots of this are $r_1,r_2,$ the closed form is $d_n = \alpha r_1^n + \beta r_2^n,$ where alpha and beta are constants. This pretty well known, but the intuition is that we expect $d_n,$ a second degree recurrence relation, to grow close to exponentially.

Anyways, we use this to find $r_1$ and $r_2$ (quadratic formula suffices).

Now, we can solve for the constants $\alpha$ and $\beta.$

This is the general method to finding the closed form of nasty second degree recurrence relations, I'm not sure if that helps you much. The main idea is the close to exponential growth of relations like these, to put it shortly.


问:
$p^p - 1$ has a prime factor of the form $lp+1$
Prove that if $p$ is a prime $p^p - 1$ has a prime factor $\equiv1(\mod p)$.
My approach was to write the congruence $p^p\equiv 1\pmod q$$(1)$ for some prime factor of $p^p-1$ $q$ and I noticed that $gcd(p,q) = 1$. Let $o =$ order $p$ of a modulo $m$. From $(1)$ $o\in\{1,p\}$. And here I don't know how to proceed. I see that it sufficies to prove that there exist a prime $q$ for which $o = p$ and we know that $o = p$ divides $\phi(q) = q-1$ so $q\equiv 1\pmod p$. Any help appreciated.

答:
As $p^p\equiv1\pmod q$

and $p^{q-1}\equiv1\pmod q$

$p^{(q-1,p)}\equiv1\pmod q$

Now if $p|(q-1),(p,q-1)=p$

Else $(p,q-1)=1,p^1\equiv1\pmod q$


问:
Fully invariant subgroups of the group $S_4$
I'm trying to find all fully invariant subgroups of the group $S_4$.
Meaning subgroups $H \leq S_4$, such that for all $\phi \in \operatorname{End}(S_4)$ the following holds: $\phi (H) \subseteq H$.

Of course the trivial subgroups $\{ e \} ,S_4$ are fully invariant, so I'm just interested in non-trivial subgroups.

I know that only characteristic subgroups (invariant under automorphisms) can be fully invariant. As $A_4, V_4$ are the only normal subgroups (invariant under inner automorphisms) these are the only possible subgroups that could be characteristic. That they are characteristic can be seen using the fact that being a characteristic subgroup is a transitive property and that automorphisms preserve order of elements. So $A_4$ must be characteristic in $S_4$, as it is generated by all elements of order $3$. By the same argument $V_4$ is characteristic in $A_4$, so in $S_4$.

Now I have to find out whether $A_4$ and $V_4$ are fully invariant. I'm missing arguments there, because I don't know many properties of endomorphisms on symmetric groups. Maybe someone has an idea?

Thanks!

答:
Here is the solution(just the comments above in one answer):

We know, both $A_4$ and $V_4$ are characteristic subgroups.

First I show, that $A_4$ is fully invariant. We know that the order of the image divides the order of the preimage. As $A_4$ is generated by all elements of order 3, the image of these generators must have order 3 or 1. In both cases it is in $A_4$.

For showing that $V_4$ is fully-invariant, we look at an endomorphism which is not an automorphism. It follows that the kernel is non-trivial, as it is normal it must be $S_4$, $A_4$, $V_4$. In all cases $V_4$ is contained in the kernel, though mapped to $e \in V_4$.


问:
Exercise on Lebesgue measure ( Treatise of Analysis Vol2 by Dieudonné)
Someone challenge me to bring the solution from anywhere! So I have posted here and see, I am optimist because this website is excellent and its members are so helpful.

Let me start with this definition:

A measure $\mu$ on a space $X$ is called diffuse if $\mu({x})=0$.

Remark [Problem 7(d) & 13.9]: can be see in this part of the book

This is the exercise: P. 211

I have captured a picture scaring of misquoting the context:

答:
Hints (assuming you have a) already, and that $X$ is metrisable and hence Polish, as the hint for c) suggests):

For b), you want to have a map $f\colon X\to [0,1]$ such that $f^{-1}[[0,t)]={U_t}$, for all $t\in[0,1]$. What would $f^{-1}[\{t\}]$ be in this case? To show that $f$ is continuous, use the fact that $\bigcap_{t'>t}U_{t'}=\bigcap_{t'>t}\overline U_{t'}$ (so it's a closed set).

Notice that the continuous function in b) would be a homeomorphism if only it was injective (as both spaces are compact). We want to make a similar construction which will result in a homeomorphism at the expense of removing a null set.

To to this, we want to find a tree $T$ of open subsets of $X$ such that for any node $\sigma$ in the tree, there are finitely many sons which are disjoint open sets contained in $\sigma$ along with their closures, and which have collectively full measure in $\sigma$. Then you can recursively match each level of the tree to a finite family of disjoint open intervals in $[0,1]$, and the matching will induce the homeomorphism.


问:
How to solve the following differential equation: $(2x^3-x^2-y)dx-(2x^2y-x)dy=0$
I'd like your help with this differential equation:
$(2x^3-x^2-y)dx-(2x^2y-x)dy=0$

It seems not difficult, but I have no idea about it

Thanks a lot for help)

答: Hint. Divide the equation by $x^2$ to get$$(2x-1)dx-2ydy+\left(\frac{dy}x-\frac{ydx}{x^2}\right)=0$$The last term is an exact differential, $d(y/x)$.


问:
About the convergence of the series $\sum_{k\ge1}\frac1{m_kp_k+1}$
As usual, let it be $p_k$ the $k$-th prime.

For each $p_k$ one can define
$$m_k=\min\,\{m\in N:\,mp_k+1\;\;is\;prime\}$$
and consider the following series
$$\sum_{k\ge1}\frac1{m_kp_k+1}\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
The series (1) seems to converge but I am not sure this is true. Up to the $10^7$-th prime, the summation gives the value $1.0278802480162903...$

Could it be found a formal way to decide the convergence of such series?

Many thanks.

答:
The random model for the primes predicts that the $m_k$ are approximatively independent random variables with $$Pr[m_k <\log^2\log p_k]\approx \sum_{d=0}^{\log^2 \log p_k}(1- \frac1{\log p_k})^d \frac1{\log p_k}$$ $$\approx 1-(1- \frac1{\log p_k})^{\log^2 \log p_k}\approx \frac{\log^2 \log p_k}{\log p_k}$$
so that $$\Bbb{E}[\sum_k \frac1{m_kp_k+1}]\le \sum_k (\frac{C\log^2 \log p_k}{\log p_k}\frac1{p_k+1}+\frac1{p_k \log^2 \log p_k+1})<\infty$$ which means that $$Pr[\sum_k \frac1{m_kp_k+1}=\infty]=0$$

It is probably out of reach to prove it does really converge. If someone can prove it diverges  they will earn the Fields medal.


答: First work modulo $p-1$ so $N:=\sum_{j=0}^{p-1}p^j=p=1$, and fix a prime factor $q$ of $N\ge3$ (and hence of $p^p-1$), which won't be $p$ or a prime factor of $p-1$. Now work modulo $q$: since $p\ne1$ but by Fermat's little theorem $p^{\gcd\{p,\,q-1\}}=1$, the GCD is $p$ as required.

Title: $p^p - 1$ has a prime factor of the form $lp+1$

Body:
Prove that if $p$ is a prime $p^p - 1$ has a prime factor $\equiv1(\mod p)$.
My approach was to write the congruence $p^p\equiv 1\pmod q$$(1)$ for some prime factor of $p^p-1$ $q$ and I noticed that $gcd(p,q) = 1$. Let $o =$ order $p$ of a modulo $m$. From $(1)$ $o\in\{1,p\}$. And here I don't know how to proceed. I see that it sufficies to prove that there exist a prime $q$ for which $o = p$ and we know that $o = p$ divides $\phi(q) = q-1$ so $q\equiv 1\pmod p$. Any help appreciated.


答:
Let $W:=(Y,Z)$, then as $Y$ and $Z$ are independent $f_W=f_Y\cdot f_Z$, therefore for $A_c:=\{(s,t)\in \mathbb{R}^2 :s/t\leqslant c\}$ we have that

$$
\begin{align*}
\Pr [Z/Y\leqslant c]&=\Pr [W\in A_c]\\&=\int_{\mathbb{R}^2}\mathbf{1}_{A_c}(s,t)f_Y(s)f_Z(t)\mathop{}\!d(s,t)\\
&=\frac1{2\pi }\left(\int_{(0,\infty )}e^{-t^2/2}\int_{(-\infty ,tc]}e^{-s^2/2}\mathop{}\!d s \mathop{}\!d t+\int_{(-\infty ,0)}e^{-t^2/2}\int_{[tc,\infty )}e^{-s^2/2}\mathop{}\!d s \mathop{}\!d t\right)
\end{align*}
$$
Differentiating respect to $c$ gives
$$
f_{Z/Y}(c)=\frac1{2\pi}\left(\int_{(0,\infty )}te^{-\frac1{2}(1+c)^2t^2}\mathop{}\!d t-\int_{(-\infty ,0)}te^{-\frac1{2}(1+c)^2t^2}\mathop{}\!d t\right)\\
=\frac1{\pi }\int_{(0,\infty )}te^{-\frac1{2}(1+c)^2t^2}\mathop{}\!d t=\frac1{\pi(1+c^2)}
$$
∎

EDIT: alternatively, using the change to polar coordinates $(t,s)=(r\cos \alpha ,r\sin \alpha )$ we have that the condition $s/t\leqslant c$ is equivalent to $\tan \alpha \leqslant c \Rightarrow  -\pi/2<\alpha \leqslant \arctan c$, and as the tangent have two full periods on $(-\pi,\pi)$ then

$$
\int_{\mathbb{R}^2}\mathbf{1}_{A_c}(s,t)f_Y(s)f_Z(t)\mathop{}\!d(s,t)=\frac1{\pi }\int_{[0,\infty )\times (-\pi/2,\arctan c]}re^{-r^2/2}\mathop{}\!d (r,\alpha )
=\frac1{\pi}(\arctan c+\frac{\pi}{2})
$$

where the result follows again by differentiation.

Title: Distribution of a division of two absolutely continuous random variables

Body:
Assume $Z,Y $ are independent standard normal random variables.

Find the distribution of $ Z/Y $.

The answer is that $ Z/Y $ is absolutely continuous random variable with density

$$ f_{\frac{Z}{Y}}(x)=\frac{1}{\pi(1+x^2)} $$

Here's what I don't understand:

I tried to write:

\begin{align}
F_{\frac{Z}{Y}}(t) & =\mathbb{P}\left(\frac{Z}{Y}\leq t\right)=\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y<0\right)+\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y>0\right) \\[8pt]
& =\mathbb{P} (Z\geq tY\cap Y<0)+\mathbb{P}\left(Z\leq tY\cap Y>0\right) \\[8pt]
& =\mathbb{P}(Z\geq tY) \mathbb{P}(Y<0)+\mathbb{P}(Z\leq tY) \mathbb{P}(Y>0) \\[8pt]
& =\frac{1}{2}\mathbb{P}(Z\geq tY)+\frac{1}{2}\mathbb{P} (Z\leq tY)
\end{align}

And then to calculate each term. for example :

$$ \mathbb{P}(Z\leq tY)=\mathbb{P} ((Z,Y)\in \{ (z,y):z\leq ty \} ) = \intop_{-\infty}^\infty \intop_{-\infty}^{ty} \frac{1}{2\pi} e^{-\frac{z^2+y^2}{2}} \, dz\, dy $$

But this leads me to a wrong answer, and I'm pretty sure that the problem is somewhere in what I wrote before.

Any help would be appreciated, thanks in advance.


问:
What is the explicit formula for the staircase function $S(x)$ defined in terms of the zeta zeros?
Riemann and von Mangoldt derived explicit formulas for the Riemann prime power counting function $J(x)$ and the second Chebyshev function $\psi(x)$ respectively via the following relationships.

(1) $\quad J(x)=\frac{1}{2\ \pi\ i}\int_{a-\infty\ i}^{a+\infty\ i}\log\zeta(s)\,\frac{x^s}{s}\ ds=li(x)-\sum _\rho Ei\left(\log(x)\ \rho\right)-\log (2)+\int_x^{\infty } \frac{1}{t \left(t^2-1\right) \log (t)} \, dt$

(2) $\quad\psi(x)=\frac{1}{2\ \pi\ i}\int_{a-\infty\ i}^{a+\infty\ i}\left(−\frac{\zeta′(s)}{\zeta(s)}\right)\frac{x^s}{s}\ ds=x-\sum_\rho\frac{x^\rho}{\rho}-\log(2\ \pi)-\frac{1}{2}\log(1-x^{-2})$

Note that the integral in (1) and (2) above also applies to the staircase function $S(x)$ as illustrated in (3) below which seems to imply the possible existence of an explicit formula for the staircase function expressed in terms of the zeta zeros.

(3) $\quad S(x)=\frac{1}{2\ \pi\ i}\int_{a-\infty\ i}^{a+\infty\ i}\zeta(s)\,\frac{x^s}{s}\ ds$

Question 1: Is it possible to derive an explicit formula for the staircase function $S(x)$ defined in terms of the zeta zeros?

Question 2: Assuming the answer to question 1 above is yes, what is the explicit formula for the staircase function $S(x)$ defined in terms of the zeta zeros?

答:
The whole number staircase function (sometimes called the integer floor function, or natural number counting function) may be expressed as,
$$
S(w) = \int_{r=1^-}^{w} \exp^*(dJ(e^v))[\ln r] dr,
$$
where $\ln r$ is the independent variable of the convolution exponential, and the argument of the c.e. is the usual prime-power staircase function (steps of $1/k$ at each $p^k$, $p$ prime; see eg "Riemann's Zeta Function," H.M. Edwards),
$$
dJ(e^u) = \left( \frac{1}{u} - \sum_{\rho} \frac{e^{(\rho-1)u}}{u} - \frac{1}{e^{u}(e^{2u}-1)u} \right),
$$
where the $\rho$ are the non-trivial zeros of the Riemann zeta function.

(Note, for those unfamiliar the convolution exponential (when it converges) is defined as
$$
\exp^*(f) = \sum_{k=0}^{\infty} \frac{f^{*k}}{k!} = \delta_0 + f + f*f/2! + f*f*f/3!+...,
$$
where $\delta_0$ is the Dirac delta function. It has some of the expected properties, like for functions $f$ and $g$ under convergence conditions, $\exp^*(f+g) = \exp^*(f)*\exp^*(g)$, and if $0<f<g$, then $\exp^*(f)<\exp^*(g)$.)

For proof see this question.


问:
Solve Linear Least Squares with $ {L}_{1} $ Norm Regularization with Linear Equality and Non Negativity Constraints
I'm searching for a way to solve the following problem in MATLAB.

$$\begin{array}{ll} \text{minimize} & \|Ax-b\|_2 + \lambda \|Cx\|_1\\ \text{subject to} & D x = 0\\ & -x \leq 0\end{array}$$

In MATLAB, lsqlin doesn't take the $\ell_1$ regularization into account, and l1_ls is only considering the constraint $x \geq 0$. Is there a way to solve the problem in MATLAB?

答:
The problem is given by:

$$\begin{aligned}
\arg \min_{x} \quad & \frac{1}{2} {\left\| A x - b \right\|}_{2}^{2} + \lambda {\left\| C x \right\|}_{1} \\
\text{subject to} \quad & D x = 0 \\
& x \geq 0
\end{aligned}$$

Probably the fastest solver to this kind of a problem is by some adaptation of the ADMM algorithm.

Yet, a simple solver can be done by the Projected Sub Gradient Method.

In this case we need a projection onto the convex set $ \mathcal{D} = \left\{ x \mid D x = 0, x \geq 0 \right\} $. Since the set $ \left\{ x \mid x \geq 0 \right\} $ isn't a sub space we can't use the alternating projection method (See Projections onto Convex Sets). So we'll use the Dykstra's Projection Algorithm from Orthogonal Projection onto the Intersection of Convex Sets.

The projection onto each set are given by (You may have a look at Projection of $ z $ onto the Affine Half Space $ \left\{ x \mid A x = b, \; x \geq 0 \right\} $):

$ \operatorname{Proj}_{ \left\{ x \mid D x = 0 \right\} } \left( y \right) = y - {D}^{T} {\left( D {D}^{T} \right)}^{-1} \left( D y \right) $.

$ \operatorname{Proj}_{ \left\{ x \mid x \geq 0 \right\} } \left( y \right) = \max \left( y, 0 \right) $.

The Sub Gradient of the objective function is given by $ {A}^{T} \left( A x - b \right) + \lambda {C}^{T} \operatorname{sign} \left( D x \right) $.

Throw all of this into a simple iterative Sub Gradient and you have a solver.

The MATLAB code is available at my StackExchange Mathematics Q3892375 GitHub Repository.

P. S.

By generating all matrices randomly the solution is usually given by the zero vector.


答:
$$(2x^3-x^2-y)dx-(2x^2y-x)dy=0$$
If you rearrange terms the DE becomes:
$$x^2(2x-1)dx+(xdy-ydx)-x^2(2ydy)=0$$
$$(2x-1)dx+d\left (\dfrac yx \right) -dy^2=0$$
Integrate.

Title: How to solve the following differential equation: $(2x^3-x^2-y)dx-(2x^2y-x)dy=0$

Body:
I'd like your help with this differential equation:
$(2x^3-x^2-y)dx-(2x^2y-x)dy=0$

It seems not difficult, but I have no idea about it

Thanks a lot for help)


问:
Can the cube of every perfect number be written as the sum of three cubes?
I found an amazing conjecture: the cube of every perfect number can be written as the sum of three positive  cubes. The equation is
$$x^3+y^3+z^3=\sigma^3$$
where $\sigma$ is a perfect number
(well it holds good for the first three perfect numbers that is):

$${ 3 }^{ 3 }+{ 4 }^{ 3 }+{ 5 }^{ 3 }={ 6 }^{ 3 }$$
$${ 21 }^{ 3 }+{ 18 }^{ 3 }+{ 19 }^{ 3 }={ 28 }^{ 3 }$$
$${ 495 }^{ 3 }+{ 82 }^{ 3 }+{ 57 }^{ 3 }={ 496 }^{ 3 }$$

Is this what I am proposing that the cube of any perfect number can be expressed in terms of the sum of three positive cubes true?

If it is then can we prove it?

答:
Well, I played myself with this formula and proved that if the following ratio holds:

$$\frac{a^3 + b^3 + c^3}{abc} = 6$$

then the three integers:

$$x_1 = a + b$$
$$x_2 = a + c$$
$$x_3 = b + c$$

are such that: $$ x_1^3 + x_2^3 + x_3^3 = y^3 $$

Proof is very simple:

Compute: $$ (a+b+c)^3$$

this gives: $$a^3 + b^3 + c^3 + 3[a^2(b+c) + b^2(a+c) + c^2(a+b)] + 6abc$$
now gather these addendums in three variables, forgetting about "6abc" for a while:

$$\alpha = a^3 + 3a^2b + 3ab^2$$
$$\beta = b^3 + 3b^2c + 3bc^2$$
$$\gamma = c^3 + 3c^2a + 3ca^2$$

now complete the gathering, adding to each variable a fraction of "6abc", obtaining:

$$x_1^3 = a^3 + 3a^2b + 3ab^2 + k_1abc$$
$$x_2^3 = b^3 + 3b^2c + 3bc^2 + k_2abc$$
$$x_3^3 = c^3 + 3c^2a + 3ca^2 + k_3abc$$

these cubes are perfect if and only if:
$$k_1abc = b^3$$
$$k_2abc = c^3$$
$$k_3abc = a^3$$

with the additional condition: $$k_1 + k_2 + k_3 = 6$$ where $$ k_1, k_2, k_3$$ are real
Now, the following equivalence holds:

$$a^3+b^3+c^3=6abc$$

and so:

$$\frac{a^3+b^3+c^3}{abc}=6$$

when this holds, the three numbers become:
$$x_1^3 = (a+b)^3$$
$$x_2^3 = (b+c)^3$$
$$x_3^3 = (c+a)^3$$

and then we have our sum of cubes.

Aftermaths!

This procedure does not say how to find the three numbers, but indeed once you find them, it is easy to show that each group of three base numbers, multiplied by an integer "h" still gives a sum of cubes that results in a perfect cube.
As an example, say:
(a, b, c) = (1, 2, 3)
then:
$$x_1 = 1 + 2 = 3$$
$$x_2 = 2 + 3 = 5$$
$$x_3 = 3 + 1 = 4$$

but also:

$$(a, b, c) = h(1, 2, 3)$$ with $$h \in \mathbb R$$ is a solution. Proof is starightforward:

if $$\frac{a^3+b^3+c^3}{abc}=6$$ holds, then also

$$\frac{h^3a^3+h^3b^3+h^3c^3}{hahbhc}=6$$ does. In fact you can group $$h^3$$ both above and beneath and then simplify.

One last thing to mention. Of course, while this does not help us in finding the three base numbers, it tells us that, once found, the number a+b+c=n has a perfect cube. Moreover, each number m=hn, with $$h \in \mathbb N$$ is a perfect cube either.

That's it. I don't know where this has been proven, because this is only the result of my spare time calculations. I hope some of you might find it interesting enough to share opinions and ideas.
Now I'm trying the same for the more general rule:

$$\sum_{i=1}^nx_i^n = y^n$$

but it already proved to be nasty!


问:
How do I translate and rotate an Ellipse in Polar coordinates?
So I'm trying to find the best ellipse that fits with a sample data, that is an easy task if the ellipses fallow the standard form:
$$r = \frac{b}{\sqrt{- e^{2} \cos^{2}{\left(θ \right)} + 1}}$$

But when I transform my data from cartesian coordinates to polar coordinates, my data will not always be close to an ellipse as standardized as this one. I need to include a variable responsible to translation $(r_0)$ and one responsible to rotation of the axis $(\theta_0)$. Thus, by obtaining these variables, I can calculate the variational of the residual of my function $r(\theta)$ relative to them, in such a way that I optimize the residual $(data sample - approximation\ function)$. I don't know what is the "general" elliptic equation (which includes translation and rotation) in polar coordinates. I searched for this equation a lot and I tried to find it by myself but I didn't succeed. Does anyone know it?

答:
Concerning the translation you have better to deduct $x_0, y_0$ from the "points" coordinates.

For the rotation you add a $\theta _0$ to $\theta$.


问:
An Olympiad Geometry problem about finding ratios.
In the figure there is a convex quadrilateral ABCD. The lines DA and CB intersect at K, the lines AB and DC intersect at L, the lines AC and KL intersect at G, the
lines DB and KL intersect at F. Prove that $\frac{KF}{FL} = \frac{KG}{GL}$ .

I feel like this question is screaming for menelaus but i can't figure out where; so i did what any desperate person would do in this case; write out all the menelaus for all of the collinear triples. But there were 10 triples so all I got was a page full of letters from which I couldn't figure out anything.

I would highly appreciate a synthetic solution using things like Menelaus, Ceva's, Area Lemma and just basic things (nothing too fancy like complex or bary)

Thanks!

答:
If you use Menelaus, you obtain $$\frac{LG \cdot KA \cdot DC}{GK \cdot AD \cdot CL} = 1,$$ where we ignore the signs of lengths.

Similarly, Ceva yields $$\frac{LF \cdot AK \cdot  DC}{FK \cdot AD \cdot CL} = 1.$$

It follows that $\frac{LG}{GK} = \frac{LF}{FK}.$

I ignored signed length out of my haste, but this is the general method of showing the result. As mentioned by Aqua, this used frequently in projective geometry, specifically because the four points form a harmonic bundle. I urge you to search that up, if you are interested :)


问:
Projection of $ z $ onto the Affine Half Space $ \left\{ x \mid A x = b, \; x \geq 0 \right\} $
This is exactly the same question as Orthogonal Projection of $ z $ onto the Affine set $ \left\{ x \mid A x = b \right\} $ except I want to project on only a half affine space instead of a full affine space.

The parameters $A$ and $b$ of the problem are as in the linked question : from this question we know that a projection of $z$ onto the affine subspace $\{x\mid Ax=b\}$ is given by :

$$x =z-A^T (AA^T)^{-1}(Az-b)$$

( we suppose here that $A$ has more columns than rows and is full rank, as in the quoted question).

What about the projection onto the half-space ?

What if I want the projections of $z$ onto the affine half-space or the full affine space, but regarding another scalar product than the canonical one, say the one related to a bilinear symmetric definite positive endomorphism given by a matrix $P$.

答:
The problem is given by:

$$\begin{aligned}
\arg \min_{x} \quad & \frac{1}{2} {\left\| x - z \right\|}_{2}^{2} \\
\text{subject to} \quad & A x = b \\
& x \geq 0
\end{aligned}$$

I am not sure there is a closed form solution for this.

Moreover, since the set $ \left\{ x \mid x \geq 0 \right\} $ isn't a sub space we can't use the alternating projection method (See Projections onto Convex Sets).

It can be solved using the Dykstra's Projection Algorithm from Orthogonal Projection onto the Intersection of Convex Sets.

The projection onto each set are given by:

$ \operatorname{Proj}_{ \left\{ x \mid A x = b \right\} } \left( z \right) = z - {A}^{T} {\left( A {A}^{T} \right)}^{-1} \left( A z - b \right) $.

$ \operatorname{Proj}_{ \left\{ x \mid x \geq 0 \right\} } \left( z \right) = \max \left( z, 0 \right) $.

A MATLAB implementation of the projection using Dykstra's Projection Algorithm which is verifed vs. CVX is available at my StackExchange Mathematics Q3537129 GitHub Repository.


答: Seek a total derivative $df=0$ by multiplying by an Ansatz $x^ay^b$ viz.$$f_x=2x^{a+3}y^b-x^{a+2}y^b-x^ay^{b+1},\,f_y=x^{a+1}y^b-2x^{a+2}y^{b+1}$$so$$0=\frac{f_{xy}-f_{yx}}{x^{a-1}y^{b-1}}=2bx^4-bx^3-(a+b+2)xy+2(a+2)x^2y^2.$$Equating coefficients gives $a=-2,\,b=0$, so$$df=(2x-1-y/x^2)dx+(1/x-2y)dy\implies f=x^2-x+y/x-y^2-C.$$The constraint therefore reduces to $x^2-x+y/x-y^2=C$.


问:
What does this dollar sign over arrow in function mapping mean?
In a certain function mapping like this,

$x \xleftarrow{\$}  \{0,1\}^k$

(Lecture Notes on Cryptography by
 S. Goldwasser and M. Bellare, page 18)

I fail to understand what exactly does this \$ sign mean. This has been put
here without any explanation or further elaboration.

It may be very trivial or may be very silly of me asking it like this, but
I want to understand what is means.

Google search (function, dollar, arrow) has not returned any result.

http://en.wikipedia.org/wiki/List_of_mathematical_symbols does not mention anything as well.

答: It is not standard notation as far as I know. I searched for dollar signs in the document, and found that he defines the notation 45 pages later on page 63.</p>


问:
Why does a cumulative distribution have the quality that $P(X<b)=\lim_{n\to\infty}\left[P\left(X\leqslant b-\frac1n\right)\right]$?
I came across the following passage in Ross' "First Course in Probability":

If we want to compute the probability that X is strictly less than b , we can apply the continuity property to obtain:
$$P\bigl(X<b\bigr)=\lim_{n\to\infty}\left[P\left(X\leqslant b-\frac1n\right)\right].$$

The continuity Ross is referring to, is the right continuity of the cumulative function but the property he mentions seems to me to be a property of a left continuity since the sequence $b-\dfrac1n$, where $n$ goes to infinity, is an increasing sequence that converges to $b$ from left to right.
Will be grateful for any enlightening remarks on this.

答: The event $X<b$ is the union of all events $X\le b-\frac 1n$ (namely, $X<b$ if and only if there exists some $n$ such that $X\le b-\frac 1n$), hence $P(X<b)$ is the supremum of all $P(X\le b-\frac 1n)$.


问:
Shared probability function of dependent Poisson variable and Binomial variable
The number of people that enter a shop is a Poisson variable with parameter $\lambda=50$. The probability that a customer that entered the shop will buy something is $p=0.16$.

I need to calculate the probability that:

exactly 20 customers, out of all the customers that entered the shop in 2 hours, will buy something

exactly 5 customers buy something if it is known that in 1 hour 40 customers entered the shop

let $X$ be the number of customers who buy something, and let $Y$ be the number of customers that entered the shop. The shared probability function $p(x, y)$ when $x=20$ is $$p(20, y)=e^{-100}\frac{100^y}{y!}\binom{y}{20}(0.16)^{20}{0.84}^{y-20}$$
hence the required probability is $$\sum_{y=20}^{\infty}e^{-100}\frac{100^y}{y!}\binom{y}{20}(0.16)^{20}{0.84}^{y-20}=\sum_{y=20}^{\infty}e^{-100}\frac{100^y}{y!}\frac{y!}{20!(y-20)!}(\frac{0.16}{0.84})^{20}{0.84}^y=\frac{e^{-100}}{20!}(\frac{0.16}{0.84})^{20}\sum_{y=20}^{\infty}\frac{100^y \cdot 0.84^y}{(y-20)!}=\frac{e^{-100}}{20!}(\frac{0.16}{0.84})^{20}\sum_{y=0}^{\infty}\frac{84^{y+20}}{y!}=\frac{e^{-100}}{20!}(16)^{20}\sum_{y=0}^{\infty}\frac{84^y}{y!}=\frac{e^{-100}}{20!}(16)^{20}e^{84}=\frac{e^{-16}}{20!}(16)^{20}\approx 0.0559$$

2.

$$P\{X=5|Y=40\}=\frac{P\{X=5, Y=40\} }{P\{Y=40\} } = \frac{e^{-50}\frac{50^{40}}{40!}\binom{40}{5}(0.16)^{5}{0.84}^{35}}{e^{-50}\frac{50^{40}}{40!}}=\binom{40}{5}(0.16)^{5}{0.84}^{35}\approx 0.1544$$

But I've a problem settling these results with what I understand, it appears, to me, in both answers that $X$ and $Y$ are independent, while I know that they aren't.

答:
For the first question, it is not necessary to appeal to the binomial distribution.  Instead, the random number of customers who buy something is itself Poisson with rate $\lambda p$, since this is a thinned Poisson process.  However, you do not specify the time increment for the original rate $\lambda = 50$.  Is it $50$ customers visiting per hour?  Per $2$ hours?  This information is absent; therefore, I will assume this is an hourly rate.

Then, with a slight modification of your notation, let $Y(t)$ represent the number of visiting customers in $t$ hours, and $X(t)$ represent the number of customers who buy something in $t$ hours.  As I mentioned above, $$X(t) \sim \operatorname{Poisson}(\lambda p t = 8t),$$ consequently $$\Pr[X(2) = 20] = e^{-16} \frac{(16)^{20}}{20!} \approx 0.0559195.$$  Your solution basically goes through a proof that the marginal (unconditional) distribution $X(t)$ is Poisson when the conditional random variable $X(t) \mid Y(t)$ is binomial and $Y(t)$ is Poisson; i.e., your solution actually proves the Poisson thinning property for the specific numeric case you are considering.

For the second question, $$X(1) \mid Y(1) = 40 \sim \operatorname{Binomial}(n = 40, p = 0.16),$$ therefore $$\Pr[X(1) = 5 \mid Y(1) = 40] = \binom{40}{5} (0.16)^5 (1 - 0.16)^{40 - 5} \approx 0.154385.$$


答: There's no sign error. You just need to identify a hidden convergent geometric series:$$\int_0^\infty\frac{x^{v-1}e^{\beta\mu}e^{-x}dx}{1-e^{\beta\mu}e^{-x}}=\sum_{n\ge1}e^{n\beta\mu}\int_0^\infty x^{v-1}e^{-nx}dx=\Gamma(v)\sum_n\frac{e^{n\beta\mu}}{n^v}.$$Incidentally, this can be rewritten with the polylogarithm as $\Gamma(v)\operatorname{Li}_v(e^{\beta\mu})$; convergence requires $\Re v>1,\,\Im\beta\mu<0$.

Title: Is the following gamma function integral correct?

Body:
$$\int_{0}^{\infty}\frac{x^{v-1}}{e^{-\beta\mu}{e^x}-1}=\Gamma(v)\left(e^{\beta\mu}+\frac{e^{2\beta\mu}}{2^v}+\frac{e^{3\beta\mu}}{3^v}+\dotsb\right)$$

This integral is given in my statistical physics assignment and I feel that the $\mu$ in the integrand should have the opposite sign (as my results would make more sense) but I don't know how to prove this and I can't find the formula anywhere. I would love to find a source for this formula.


问:
Why doesn't the Continuum Hypothesis show that $\text{ZFC}$ is simply consistent?
Where is my error?

A system is simply inconsistent is every formula is provable (if $\text{A} \wedge\neg \text{A}$ were provable for some formula $\text{A}$, then by $\neg$-introduction, any formula can be proven).  Taking the contraposition, we get that if there exist some formula that is unprovable, the system is simply consistent.

Since the Continuum Hypothesis is undecidable and thus unprovable in $\text{ZFC}$, shouldn't that mean that $\text{ZFC}$ is simply consistent?  This also shows the other direction of Gödel's first theorem: if the system is incomplete, then it is consistent.

I know all of this can't be right because I haven't read any of this and it can't be that easy, so I must be missing something.

答:
The think you're missing is the conditional part of undecidability of CH: it is undecidable, provided ZFC is consistent.

Roughly, the proof goes like this: we start with an arbitrary model of ZFC (which we have if ZFC is consistent), and then the constructible universe L within is a model of ZFC+CH, while an appropriate Cohen forcing extension is a model of ZFC+$\neg$CH.

None of this makes any sense if we have no model of ZFC to begin with.


问:
The parametric representation of a line
I have a lecture for my linear algebra and geometry module that focuses on the cross product and line and planes within Euclidean geometry. The lecture notes talk about the parametric representation of a line which is something I covered in my A-levels, however I can't make the connection in my head between what I learned at A-level and what was in the lecture leading to me not being able to understand this content. The part that I am struggling on is as follows:

The straight line through the point $P\in\Bbb R^n$ in direction of vector $\vec a$ has the parametric form:
$$\vec x (t)=P+\vec a t\quad\forall t\in\Bbb R$$
Similarly, given two points $P, Q\in\Bbb R^n$, the line passing through $P$ and $Q$ has parametric form:
$$\vec x (t)=P+t(Q-P)\quad\forall t\in\Bbb R$$
I was wondering if anyone could explain how this links back to earlier content that I have studied, hopefully allowing me to understand this?

答:
$
\newcommand{\vect}[1]{\vec{\boldsymbol{\mathbf{{#1}}}}}
\newcommand{\R}{\mathbb{R}}
$
A good place to start would be a visual example. Suppose $\vect a = (1,2)$ and $\vect y_0= (-1,0)$. We want the line through $\vect y_0$ in the direction of $\vect a$. This is given by

$$\left\{ \vect x \in \Bbb R^2 \; \middle| \; \vect x = t \vect a + \vect y_0 \text{ for some } t \in \Bbb R \right\}$$

Here is what we get if we visualize the elements of this set for $t \in \{-3,-2,-1,0,1,2,3\}$:

What do we notice? This appears to be a line with slope $2$, and $y$-intercept $2$ as well. That means the equation of the line -- once we span over all of $t \in \R$ -- is given by

$$y = 2x+2$$

That's interesting, because $2 = a_y/a_x = 2/1$ (where $a_x$ denotes the $n$ component of $\vect a$, and similar for $y$). $y$ in particular goes through $\vect y_0$. So let's consider $\vect a$ as representing a slope through the ratio of its points, and $\vect y_0$ as a single point through which the line goes. This uniquely determines the line via its point slope form: the equation of a line with slope $m$ going through $(x',y')$ is given by

$$y - y' = m(x-x')$$

Given a vector $\vect a$ and $\vect y_0$, then

$$y - y_{0,y} = \frac{a_y}{a_x} (x - y_{0,x})$$

Plugging in $y_{0,x} = -1$, $y_{0,y} = 0$, $a_y = 2$, and $a_x = 1$, we have

$$y - 0 = \frac{2}{1} (x - (-1)) \implies y = 2x+2$$

We can work this similarly for your "line through two points" thing. Suppose we have, now, $\vect a = (1,3)$ and $\vect b = (2,1)$. The line in the green is the unique line through these two points:

Notice however, we can get a third vector from this (up to its orientation): the vector $\vect c = \vect b - \vect a$ (or the vector $\vect d = \vect a - \vect b$). Choosing $\vect c$ because why not, we then have this picture:

Notice, however, the line that we want is precisely the set of all vectors that are scalar multiples of $\vect c$ from that position. That is, we have the line is given by

$$\left\{ \vect x \in \Bbb R^2 \; \middle| \; \vect x = t \vect c + \vect a \text{ for some } t \in \Bbb R \right\}$$

Some Notes: You can use $\vect d$ in lieu of $\vect c$, and $\vect a$ in lieu of $\vect b$ in the above definition. We just want a line through $\vect a$ and $\vect b$, and $\vect c$ or $\vect d$ go in the direction necessary. Carefully examine how this is similar to our previous example! The choice of $\vect c$ versus $\vect d$, or $\vect a$ versus $\vect b$, is superfluous for this application and the end result is ultimately the same even if the definition looks different. If we were to limit $t$ to $0 \le t \le 1$, as we do when parameterizing a line segment specifically, it would be a different matter.

Anyhow, since $\vect c = \vect b - \vect a$, then we have that the set is equivalent to

$$\left\{ \vect x \in \Bbb R^2 \; \middle| \; \vect x = \vect a + t \left(\vect b - \vect a \right) \text{ for some } t \in \Bbb R \right\}$$

You can then re-analyze this through our opening example and explanation -- a line with given slope through a point, since $\vect b - \vect a$ (the thing $t$ scales) determines the slope of the line, and $\vect a$ determines the location through which the line goes.

Title: The parametric representation of a line


问:
Projection onto the Intersection of 2 Affine Subspaces
I was wondering if there is a closed formular for the projection onto the intersection of the subspaces $Ax = b$ and $Zx = 0$. I know there is a closed formula for either one of those, but can you also project onto the interesection by use of the pseudoinverse?

I am aware of the alternating projection method, but this takes too long for my purposes.

Thanks!

答:
The problem is given by:

$$\begin{aligned}
\arg \min_{x} \quad & \frac{1}{2} {\left\| x - z \right\|}_{2}^{2} \\
\text{subject to} \quad & A x = b \\
& Z x = 0
\end{aligned}$$

Which is equivalent of:

$$\begin{aligned}
\arg \min_{x} \quad & \frac{1}{2} {\left\| x - z \right\|}_{2}^{2} \\
\text{subject to} \quad & C x = d \\
\end{aligned}$$

Where $ C = \begin{bmatrix} A \\ Z \end{bmatrix} $ and $ d = \begin{bmatrix} b \\ 0 \end{bmatrix} $.

Then the answer is given by $ x = z - {C}^{T} {\left( C {C}^{T} \right)}^{-1} \left( C z - d \right) $.

It can be easily generalized to more than 2 sub spaces.


问:
I can't understand the result of this limit
$$ \lim_{n\to\infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3}
$$

I think the answer is $-\infty$ because I take the highest degree coefficients $$n^4  (numerator)$$ and $$-n ^ 3(denominator)$$

and I was thinking $+/- = -$, but the actual answer is $+\infty$, this is probably a dumb question, but I didn't understand why is $+\infty$ instead of $-\infty$.

答:
First of all it should either be $ n \to \infty $ or you should replace $ n $ with $ x $. It is

$ \displaystyle\lim_{x \rightarrow\infty  }\frac{x^4-x^3+1}{\sqrt{x}+x^2-x^3} = \lim_{x \rightarrow\infty  }\frac{x^4}{-x^3} = \lim_{x \rightarrow\infty  } (-x) = -\infty $


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答: The imaginary unit $i$ is just a unit, like $m$ (meters), $s$ (seconds), or $\hat{u}$ (a unit vector). Any unit can be multiplied by a scalar, so $-i$ is jus the result of multiplying the imaginary unit by $-1$, no different than $4i$, $6i$, or $2.5i$.</p>

Title: What is $-i$ exactly?