问:
Lei Fu(Etale Cohomology Theory) Proposition 3.2.5
I'm having trouble understanding the proof of Proposition 3.2.5 (i) of "Etale Cohomology Theory" by Lei Fu. The statement of this proposition is as follows:

Proposition 3.2.5. Let $X$ be a scheme on which a finite group $G$ acts admissibly on the right and let $Y=X/G$ be the quotient of $X$ by $G$. Suppose $X$ is of finite presentation over $Y$. If the inertia group $G_i(x)$ is trivial for any $x\in X$, then $X$ is etale over $Y$.

In the proof of this, the author first takes the strict henselization of $\mathcal{O}_{Y,y}$ with respect to a separable closure of $k(y)$ for arbitrary $y\in Y$, which is denoted by $\widetilde{\mathcal{O}}_{Y,\bar{y}}$, and then he mentions Proposition 2.5.10,whose statement is as follows:

Proposition 2.5.10. Consider a Cartesian diagram
$\require{AMScd}$
\begin{CD}
X'=X\times_Y Y' @>{g'}>> X\\
@V{f'}VV @V{f}VV\\
Y' @>{g}>> Y
\end{CD}
Suppose that $f:X\to Y$ is locally of finite presentation and $g:Y'\to Y$ is faithfully flat. If $f'$ is unramified (resp. etale, resp. smooth), then so is $f$.

Using this, he sais that it suffices to show that $X\times_Y\mathrm{Spec}\,\widetilde{\mathcal{O}}_{Y,\bar{y}}$ is etale over $\mathrm{Spec}\,\widetilde{\mathcal{O}}_{Y,\bar{y}}$ and reduces to the case where $Y$ is a spectrum of a strictly henselian local ring. What I don't understand is this point. Generaly, the map $\mathrm{Spec}\,\widetilde{\mathcal{O}}_{Y,\bar{y}}\to Y$ is not faithfully flat, so I tried to reduce to the case in where $Y$ is a local scheme by replacing $X$ by $\coprod_{g\in G} \mathcal{O}_{X,x^g}$(on which $G$ acts on the right) and $Y$ by $\mathrm{Spec}\,\mathcal{O}_{Y,y}$. However I can't prove that the quotient of $\coprod_{g\in G} \mathcal{O}_{X,x^g}$ by $G$ is equal to $\mathrm{Spec}\,\mathcal{O}_{Y,y}$ (or perhaps is this a wrong statement?). If this statement is wrong, what is the right approach to proving this proposition?

Sorry for my poor English, and I really appreciate your help.

答: A morphism $f:X \to Y$ is étale if and only if it for all $y \in Y$ the basechange of $f$ along $\operatorname{Spec} \mathcal{O}_{Y,y}$ is étale. Lets take this for granted (c.f. https://stacks.math.columbia.edu/tag/02GU); the proof depends on your definition of étale. Your proposition 2.5.10 tells us that we can check étaleness after further basechange to $\overline{\operatorname{Spec}} \mathcal{O}_{Y,\overline{y}}$.


问:
Induction for binomial coefficients
I would like some help to prove the following equality :
$$\sum_{i=0}^n \binom{n}i^2=\binom{2n}n$$
I wanted to do a proof by induction :
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=1}^{n+1} \binom{n+1}i^2=1 + \sum_{i=0}^{n} \binom{n+1}{i+1}^2=1+\sum_{i=0}^{n} \bigg(\binom{n+1}i+\binom{n}{i+1}\bigg)^2$$
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=0}^{n} \bigg(\binom{n}i^2+2\binom{n}i\binom{n}{i+1}+\binom{n}{i+1}^2\bigg)=1+\binom{2n}n+ \sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1} +(\sum_{i=0}^n \binom{n}i^2-1) $$
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=2\binom{2n}n+\sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1}=2\bigg(\sum_{i=0}^n \binom{n}i^2(1+\frac{n-i}{i+1})\bigg)=2(n+1)\bigg(\sum_{i=0}^n \binom{n}i^2\frac{1}{i+1}\bigg)$$

But now I'm stuck.

答:
The binomial theorem says $$(1+x)^n=\sum^n_{i=0}\binom{n}{i}x^i,$$ and we know that
$$(1+x)^n(1+x)^n=(1+x)^{2n}.$$ Comparing the coefficient of $x^n$, we get $$\sum^n_{i=0}\binom{n}{i}\binom{n}{n-i}=\sum^n_{i=0}\binom{n}{i}^2$$ on the LHS, and $\binom{2n}{n}$ on the RHS.

ContentLicense: CC BY-SA 4.0

Title: Induction for binomial coefficients


问:
probability of multivariate
given a joint cdf
$$ \dfrac{1}{1+e^{-x}}\dfrac{1}{1+e^{-y}}$$
where $x\in \mathbb{R}$ and $y\in \mathbb{R}$. Find $P(X \geq 0, Y\geq 0)$. I take the integral and find that it diverges but the notes i have ask me to find an answer (Answer with a fraction.) am i not seeing something? Thanks :)!

Y

Title: probability of multivariate

Body:
In your case $X,Y$ are independent:

$$F_{XY}=F_X F_Y$$

Thus

$$P(X>0,Y>0)=(1-F_X(0))(1-F_Y(0))=1/4$$


答: The expression is the cdf, NOT pdf.  You don't integrate.  The probability you want is one - probability that either is negative $=(1-\frac{1}{1+e^0})^2=\frac{1}{4}$.

Title: probability of multivariate

Body:
given a joint cdf
$$ \dfrac{1}{1+e^{-x}}\dfrac{1}{1+e^{-y}}$$
where $x\in \mathbb{R}$ and $y\in \mathbb{R}$. Find $P(X \geq 0, Y\geq 0)$. I take the integral and find that it diverges but the notes i have ask me to find an answer (Answer with a fraction.) am i not seeing something? Thanks :)!

Y


问:
demonstrate the following argument
argument to be demonstrated:

$A ∨ (B ∧ C)$

$B → D$

$C → E$

$D ∧ E → A ∨ C$

$¬A$

$∴ C$

My attempt at this mathematical demonstration was as follows:

$A ∨ (B ∧ C)$

$≡ (A ∨ B) ∧ (A ∨ C)$

$≡ (A ∨ A) ∧ (B ∨ C)$

$≡ A ∧ (B ∨ C)$

$≡ (A ∧ B) ∨ (A ∧ C)$

答:
In your attempt, the 3rd line is incorrect. If A is false then $A\lor (B\land C) $ (1st line) is equivalent to $B\land C.$ But if $A$ is false then $A\lor (B\lor C) $ (3rd line)  is equivalent to $B\lor C.$ Also, you must incorporate some of the given premises into the argument  to conclude that $C$ is true. Actually,  the first premise $A\lor (B\land C) $ and the fifth premise $\neg A$ are sufficient.

Rule: If $\neg X$ then $(X\lor Y)\implies Y.$

Rule: $\neg ((\neg A)\land A).$

Applying these when  $X$ is $((\neg A)\land A)$ and when  $Y$ is $(\neg A)\land (B\land C),$ we have $$[\,(\neg A)\land (A\lor (B\land C))\,]\implies [\,((\neg A)\land A))\,\lor\, ((\neg A)\land (B\land C))\,]\implies $$ $$\implies [\,(\neg A)\land (B\land C)\,]\implies$$ $$\implies  [\,(B\land C)\,]\implies C$$

Title: demonstrate the following argument


问:
How to prove that $f(z) := \bar{z}$, for $z \in \mathbb{C}$, is bounded on a rectangle $\mathcal{R} \subseteq \mathbb{C}$?
Let $f(z):=\bar{z}$, for $z=x+iy \in \mathbb{C}$. Explain why $f(z)$ is bounded on the rectangle $$\mathcal{R} := [a,b] + i[c,d] = \{z \in \mathbb{C} : a \leq \operatorname{Re}(z) \leq b , \hspace{0.2em} c \leq \operatorname{Im}(z) \leq d \}$$ with $a<b \in \mathbb{R}$ and $c<d \in \mathbb{R}$.

My attempt:

I proved this using the definition of a bounded function, i.e., if $f(z)$ is bounded, then there exists $M \geq 0$ such that $$M := \sup_{z \in U} \lvert f(z) \rvert < \infty$$ for $U \subseteq \mathbb{C}$.

In this case, I showed that
\begin{align*} M &= \sup_{z \in \mathcal{R}} \lvert f(z) \rvert = \sup_{z \in \mathcal{R}} \lvert \bar{z}^{2} \rvert = \sup_{x \in [a,b], \\ y \in [c,d]} \lvert x-iy \rvert^{2} \leq \\ &\leq \sup_{x \in [a,b], \\ y \in [c,d]} \lvert x+iy \rvert^{2} = \sup_{z \in \mathcal{R}} \lvert \operatorname{Re}(z) + i\operatorname{Im}(z) \rvert^{2} \\ &= \lvert b +id \rvert^{2} < \infty.\end{align*}
Therefore, $f(z)$ is bounded on $\mathcal{R}$. However, I'm not sure what kind of explanation (without proof) I can give to justify my calculations.

Since $f(z)$ is continuous and defined for all $z \in \mathbb{C}$, and $\mathcal{R}$ is closed (and therefore bounded), can I say that $f(\mathcal{R})$ is compact? and by saying this, would it be enough to justify the fact that $f(z)$ is bounded?

答:
Without loss of generality, take $|a|<|b|$ and $|c| <|d|$.

Now $|f(z)|^2 = |z|^2 \le b^2 + d^2$ is bounded on the rectangle.


答: Thistlethwaite's algorithm uses coset spaces, not factor groups. The groups aren't normal. The function is IsNormal(G, H), i.e. the bigger group goes first, but you're plugging in the smaller group first. As GroupProps notes, "Note that if the order of the group and the subgroup are interchanged, GAP will not detect an error and will return true, because any subgroup of a group normalizes the whole group."

Title: GAP: Rubik's Cube Group - Thistlethwaite's algorithm

Body:
In GAP let $G_0$ be the Rubik's Cube Group defined by the six moves

```<code>U := (1,3,8,6)(2,5,7,4)(9,33,25,17)(10,34,26,18)(11,35,27,19);
L := (9,11,16,14)(10,13,15,12)(1,17,41,40)(4,20,44,37)(6,22,46,35);
F := (17,19,24,22)(18,21,23,20)(6,25,43,16)(7,28,42,13)(8,30,41,11);
R := (25,27,32,30)(26,29,31,28)(3,38,43,19)(5,36,45,21)(8,33,48,24);
B := (33,35,40,38)(34,37,39,36)(3,9,46,32)(2,12,47,29)(1,14,48,27);
D := (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40);
G0 := Group([U, L, F, R, B, D]);
</code>```

Then define the subgroups, see Thistlethwaite's Algorithm

```<code>G1 := Subgroup(G0, [L,   R,   F,   B,   U^2, D^2]);
G2 := Subgroup(G0, [L,   R,   F^2, B^2, U^2, D^2]);
G3 := Subgroup(G0, [L^2, R^2, F^2, B^2, U^2, D^2]);
</code>```

I try to define quotient groups with GAP, but unfortunately I get the following error:

```<code>FactorGroup(G2, G3);
Error, <N> must be a normal subgroup of <G> at /proc/cygdrive/C/gap-4.11.0/lib/grp.gi:2569 called from
<function "FactorGroup">( <arguments> )
 called from read-eval loop at *stdin*:13
you can 'quit;' to quit to outer loop, or
you can 'return;' to continue
</code>```

I do not understand this error since

```<code>IsNormal(G3, G2);
true
</code>```

tells me that $G_3$ is indeed normal in $G_2$. What is wrong with my code?


问:
Determine a possible quartic polynomial equation such that $f(x) > 0$ for $-4 < x < -2$ and $3 < x < 7$
In typical high school fashion, nowhere in the curriculum was there a question about creating a polynomial equation. Yet here it is in the exam practice questions. Little help please?</p>

答:
A polynomial will change sign at the roots.

If the roots are $-2,-4, 3, 7$ then $P(x) = a(x+2)(x+4)(x-3)(x-7)$ and for $x\le 3$ we have $x+2,x+4, x-3, x-7 < 0$ and $P(x) \le 0$ so $a < 0$ and for $-2 < x < -4$ we have $(x+2) > 0, x+4 < 0, x-3< 0, x-7 < 0$ so $a < 0$ means $P(x) > 0$ and so on.

SO $P(x) = -(x+4)(x+2)(x-3)(x-7)$ is one possible solution.

But the question never said $f(x) \le 0$ outside the intervals.  So long as the roots (if any) are outside the intervals and there are the right even or odd number of roots between $-2$ and $3$ it will work.

$P(x) = (x+1)x(x-1)(x-2)$ will be acceptable as $x+1,x,x-1, x-2 < 1$ for $-4 < x < -2$ so $P(x) > 0$ there and $x+1,x, x-1, x-2 > 0$ for $3< x< 7$ so $P(x) > 0$ there.

We can even do $P(x) = x^4 + 1$ as that is positive anywhere. (and has no roots).


问:
Proving a and b are perfect squares if and only if $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$ are perfect squares

Let $a$, $b$ be two positive integers. Prove that $a$ and $b$ are both perfect squares if and only if both $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$ are perfect squares.

I believe the proof is based in prime factorization, as if $a$ is a perfect square, then $a = (p_1^{a_1}\cdot\ldots\cdot p_n^{a_n})^2$, and the same goes for the gcd and lcm when they are perfect squares. I'm just a little lost on how to connect the dots and prove both directions.

答: Hint: For any given prime $p$, let the exponent of $p$ in the prime factorization of $a$ be $a_p$ and let the exponent of $p$ in the prime factorization of $b$ be $b_p$ (where $0$ is allowed as an exponent in both cases). In terms of $a_p$ and $b_p$, what are the exponents of $p$ in $\operatorname{lcm}(a, b)$ and in $\gcd(a, b)$ respectively?


问:
Orienting curves with differential forms
Consider the circle given by the equation $x^2+y^2=1$. We can orient this curve by choosing the tangent vector field $(-y,x)^T$, which defines a direction.

Supposedly we can do this with by taking an appropriate differential form $\omega$. How is this done? The idea, I thought, should be to find a 1-form such that $\omega((-y,x)^T)>0$ for all $x,y$, but this seems to be impossible since all 1-forms look like $adx+bdy$.

EDIT: If we are allowed to orient curves with form fields, then this isn't too bad! Is this allowable?

答:
First, a remark about "form" vs "form field": as Hurkyl said, the words "differential form", or "differential $k$-form", or simply "$k$-form" usually refer to a smooth choice of a covector at every point. Differential geometers also say "tensor" when they mean "tensor field".

Second, a choice of a top-dimensional nowhere vanishing differential form is what is usually understood as orientation: to orient an $n$-manifold is to specify a nowhere vanishing $n$-form. Any two such forms are related by multiplication by a nowhere vanishing smooth function (hence, either positive or negative). The equivalence class up to multiplication by positive functions constitutes an orientation.

Finally, the construction of nowhere vanishing $1$-form on a circle was discussed here.

ContentLicense: CC BY-SA 3.0


问:
Is there an example of a linear map between Banach Spaces that is not closed?
Is there a simple example of a linear map $T: X \to Y$ between Banach spaces that is not closed, i.e. if $x_n \to 0$ in $X$ and $T(x_n) \to y$ in $Y$, then $y=0$? I know the Closed Graph Theorem says that if $T: X \to Y$ is a linear map between Banach spaces, then
\begin{align*}
T \text{ is bounded } \iff T \text{ is closed } \iff \text{gr}(T) \text { is a closed subspace of } X \oplus Y.
\end{align*}
So I would have to find an unbounded linear map between Banach spaces, correct? What exactly would that be?

答:
We can construct an unbounded linear functional easily. Pick your favorite basis $e_n$ with $\lVert e_n \rVert = 1$ of $X$ (which can be done with the axiom of  choice) and define $f$ as
$$f(x)=\sum_n n x_n,\,\,\,\,\text{where }\,\,\,\,x=\sum_n x_n e_n.$$
Note that this is well-defined, as $x_n$ is non-zero for finitely many $n$ for every $x$. Then pick
$$y_n = \frac{e_n}{\sqrt{n}},$$ which clearly goes to $0$, but $f(y_n)=\sqrt{n} \to +\infty$.


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

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

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

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

答: The divergence of the Harmonic Series.</p>

LastEditorDisplayName: anon

Body: The divergence of the Harmonic Series.


问:
Why does $v \in \ker(A- \lambda I)^m$ imply that $Av$ is contained in $\ker(A- \lambda I)^m$?
Given an $n$ dimensional vector space $V$ and an $n \times n$ matrix $A$,

Then why does $v \in \ker(A- \lambda I)^m$ imply that $Av$ is contained in  $\ker(A- \lambda I)^m$?

I'm not seeing it.

答: This is essentially because $(A-\lambda I)^{m}$ and $A$ commute. Can you see this now?</p>


答:
The question only makes sense if $f(x)>0$ only in the given range. Then by continuity, $f(x)$ must be $0$ at $x=-4,-2,3,7$, and we take

$$f(x)=\pm(x+4)(x+2)(x-3)(x-7),$$ which is indeed a quartic polynomial.

The sign alternates between the roots so taking the $-$ sign fulfills the request.

Title: Determine a possible quartic polynomial equation such that $f(x) > 0$ for $-4 < x < -2$ and $3 < x < 7$

Body: In typical high school fashion, nowhere in the curriculum was there a question about creating a polynomial equation. Yet here it is in the exam practice questions. Little help please?</p>


问:
Integration and step functions, related to Rudin Theorem 6.15
I'd like any advice I could get about this problem. In general, I feel as though this particular step function has been poorly discussed in class. We are not allowed to assume continuity either, and I'm confused about how to approach this otherwise.

Let $f:[a,b] \rightarrow \mathbb{R}$ be bounded, let $a<y<b$ and $\alpha:[a,b] \rightarrow \mathbb{R}$ be $\alpha (x)=I(x-y)$.

Prove that $f$ is Riemann-Stieltjes integrable if and only if $lim_{x\rightarrow y^+}f(x)=f(y)$, and $\int_{a}^{b}fd\alpha=f(y)$.

Any help is appreciated. Thanks!

答:
Here are a couple of hints that I think show what's going on here pretty well.

$\lim_{x \to y^+} f(x) = f(y)$ will give you right continuity at $y$; notice that you don't need left continuity for the result as the integral has $ \alpha(x) = I(x - y) = 0$ for $y \geq x$.

You partition $[a, b]$ as $a = x_0< x_1 = y < x_2 < x_3 = b$. Then take the limit as $x_2 \to y$ and notice that for any partition an $[x_{n-1}, x_{n}]$ so that $x_{n-1} > y$ we have
$$
M_n \Delta \alpha(x_n) = M_n (I(x_n-y) - I({x_{n-1}-y})) = 0
$$
where $M_n$ is the $\sup$ over the interval $[x_{n-1}, x_n]$; this holds analogously  for the $m_n$, the corresponding $\inf$.


问:
What $x$ satisfies $3^{4x}\cdot 9^6 = 81^8$?

What $x$ satisfies $3^{4x}\cdot 9^6 = 81^8$?

I reckon it is possible to rearrange bases $3$ and $9$ to $81$, but I don't know how.
Which rules apply here?

Body:
$$3^{4x}\cdot 3^{12}=3^{32}$$

Thus

$$4x+12=32$$

$$x=5$$


问:
A proof that every $R$-basis of $M$ has the same number of elements.
Here is the proposition my professor gave to us:

Assume $R$ is commutative with unity.

Let $M$ be a finitely generated free $R$-module. Then every basis of $M$ has the same number of elements called the rank of $M.$

Here is the proof:

Let $\mathcal {B} = \{v_1, \dots , v_n\}$ and $\mathcal {B'} = \{w_1, \dots , w_r\}$ be two $R$-bases. There exist $P \in M_{r \times n}(R),Q \in M_{n \times r}(R)$ such that $$P\begin{bmatrix}
v_1 &\\
\vdots\\
v_n
\end{bmatrix} = \begin{bmatrix}
w_1 &\\
\vdots\\
w_r
\end{bmatrix}$$ and $$Q\begin{bmatrix}
w_1 &\\
\vdots\\
w_r
\end{bmatrix} = \begin{bmatrix}
v_1 &\\
\vdots\\
v_n
\end{bmatrix}.$$

Then $QP = I_{n\times n}$ and $PQ = I_{r\times r}.$

Case(1). (r > n)

Then $I_{r \times r} = [P_{r \times n} \mid Q_{r \times (r-n)}] [\frac{Q_{n \times r}}{Q_{(r-n)\times r}}],$ which is a contradiction since $\det [P_{r \times n} \mid Q_{r \times (r-n)}]=0.$

By symmetry, $r < n$ also can not occur. So $r = n.$

My question is that I do not understand why  $\det [P_{r \times n} \mid Q_{r \times (r-n)}]=0.$ Could anyone explain this to me please?

答:
The proof as you've written it is full of errors.

You say $I_{r \times r} = [P_{r \times n} \mid Q_{r \times (r-n)}] [\frac{Q_{n \times r}}{Q_{(r-n)\times r}}]$. Well, $Q_{r \times (r-n)}$ doesn't make sense since $Q$ is $n \times r$ and we're assuming $r > n$. You must have meant $I_{r \times r} = [P_{r \times n} \mid P_{r \times (r-n)}] [\frac{Q_{n \times r}}{Q_{(r-n)\times r}}]$. I'm guessing $P_{r \times (r-n)}$ means you pick off the first $r-n$ columns. In that case, $[P_{r \times n} \mid P_{r \times (r-n)}]$ does have determinant zero since it includes repeated columns. (This is where commutativity of $R$ is used.)

However, the corrected claim is $I_{r \times r} = [P_{r \times n} \mid P_{r \times (r-n)}] [\frac{Q_{n \times r}}{Q_{(r-n)\times r}}]$. The right-hand side is $P_{r \times n} Q_{n \times r} + P_{r \times (r-n)} Q_{(r-n) \times r}$. The first term is indeed $I_{r \times r}$, but I see no reason for the second term to be 0. What you want is to use $0_{(r-n) \times r}$ rather than $Q_{(r-n) \times r}$. Then it does work. You may as well have used $0_{r \times (r-n)}$ as well, which is entirely more transparent.


问:
Rearranging Matricies
I am little confused about re-arranging matrix equations... So I know you cannot rearrange multiplication of matricies like you would normally with algebra as you cannot divide, but can you still do something like this when multiplication is not in the question?

T+Y = D

  T = D-Y

答: Yes, matrices form a vector field under addition and multiplication by scalars. It is sometimes useful to fold matrices into vectors, e.g., but stacking their columns into one tall column vectors.  One can introduce the notions of norm and inner product in this way.</p>

Title: Rearranging Matricies


问:
Computing continued fraction expansions
My question concerns the numerical accuracy of a continued fraction expansion.  A typical algorithm for computing a continued fraction can be written in Python as :

```<code>x0 = sqrt(2)

N = 40
a = [0]*N
u = [0]*N

x = x0
for k in range(N):
    a[k] = int(x//1)
    u[k] = x % 1      # Often replaced with x - a[k]  ???
    x = 1/u[k]

print(a)
</code>```

For $x=\sqrt{2}$, this produces

```<code>[1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 1, 3, 1, 1, 2, 1809, 1, 2, 5, 2, 2, 1, 2, 1]
</code>```

Or for $x = \pi$, I get :

```<code>[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 3, 3, 23, 1, 1, 7, 4, 35, 1, 1, 1, 2, 3, 3, 3, 3, 1, 1, 14, 6, 4, 5, 1, 7, 1, 5, 1]
</code>```

One observation is that round-off error has clearly crept into the calculation.  The expansion for $\sqrt{2}$ should be repeating 2s, and the approximation for $\pi$ is accurate up to the first appearance of 14 (12 terms after the leading $a_0=3$ term).

What I also found is that while round off error does creep in, the number of correct terms in the expansion is exactly what is needed to get convergents that agree with $x$ up to machine precision.  So for example,  convergent 20 of $\sqrt{2}$  is given by

```<code>54608393 / 38613965
</code>```

This convergent approximates $\sqrt{2}$ to $4.44 \times 10^{-16}$.  Convergent 21 agrees with $\sqrt{2}$ exactly (in finite precision arithmetic).  Interestingly, the expansion has exactly 20 correct terms (i.e. 2s) after the leading $a_0 = 1$.

Similarly, for $\pi$, the convergent 12 is given by

```<code>80143857 / 25510582
</code>```

which agrees with $\pi$ to within an error of $4.44 \times 10^{-16}$.  Convergent 13 agrees with $\pi$ exactly (in finite precision arithmetic).

In both cases, the numerators for the convergents agree with the OEIS. See A001333 and A002485.

The above observations led me to these questions :

Is there a standard stopping criteria that can be used to determine when the expansion has converged to a desired tolerance?

Is it always the case that one will have enough correct terms in the expansion to approximate the desired number to machine precision?

Is it possible to detect whether a continued fraction is periodic?  Or, if one knows it will be periodic (i.e. $x$ is a root of a quadratic with integer coefficients), is it possible to get the periodic sequence exactly?

It has also occurred to me that nobody would think of computing continued fraction expansions using finite precision arithmetic!  I would like to do this problem as an exercise in a beginning computational math course, and was hoping not to go into variable precision arithmetic (an area which is really outside of my wheel house).

答:
took a while, I wrote the Gauss-Lagrange method using the left neighbor instead of the right neighbor. Given a reduced form $Ax^2 + B xy + C y^2,$  which means that $D= B^2 - 4 AC$ is positive but not a square, $AC < 0 $  and $B > |A+C|,$ the numbers listed that I call "epsilon" (well, its absolute value) give the partial quotients for the continued fraction of
$$ \frac{B+ \sqrt D}{2A}  $$
when $A$ is positive. I have not tried negative $A$ yet.. sample run

Sample: $47x^2 + 321 xy - 37 y^2$
and $D=109997,$  target is $ \frac{321+ \sqrt{ 109997}}{94}  $

First, a few "partial quotients" in the continued fraction expansion

```<code>? x = 321 + sqrt(109997)
%3 = 652.6579563345345104401576022
? x /= 94
%4 = 6.943169748239728834469761725
?
? x -= floor(x) ; x = 1/x
%5 = 1.060254531982535997122061997
? x -= floor(x) ; x = 1/x
%6 = 16.59626200880353974842520003
? x -= floor(x) ; x = 1/x
%7 = 1.677115068938572016859051397
? x -= floor(x) ; x = 1/x
%8 = 1.476853855235527410365538515
? x -= floor(x) ; x = 1/x
%9 = 2.097078568246198860000621270
? x -= floor(x) ; x = 1/x
%10 = 10.30093477958926629742181316
? x -= floor(x) ; x = 1/x
%11 = 3.322979156363579950723385592

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycleLeft  47  321  -37

0  form   47 321 -37   epsilon  6
1  form   -271 243 47   Epsilon  -1
2  form   19 299 -271   Epsilon  16
3  form   -191 309 19   Epsilon  -1
4  form   137 73 -191   Epsilon  1
5  form   -127 201 137   Epsilon  -2
6  form   31 307 -127   Epsilon  10
7  form   -97 313 31   Epsilon  -3
8  form   97 269 -97   Epsilon  3
9  form   -31 313 97   Epsilon  -10
10  form   127 307 -31   Epsilon  2
11  form   -137 201 127   Epsilon  -1
12  form   191 73 -137   Epsilon  1
13  form   -19 309 191   Epsilon  -16
14  form   271 299 -19   Epsilon  1
15  form   -47 243 271   Epsilon  -6
16  form   37 321 -47   Epsilon  8
17  form   -247 271 37   Epsilon  -1
18  form   61 223 -247   Epsilon  4
19  form   -163 265 61   Epsilon  -1
20  form   163 61 -163   Epsilon  1
21  form   -61 265 163   Epsilon  -4
22  form   247 223 -61   Epsilon  1
23  form   -37 271 247   Epsilon  -8
24  form   47 321 -37
  form   47 x^2  + 321 x y  -37 y^2

minimum was   19rep   x = -7   y = 1 disc 109997 dSqrt 331
Automorph, written on right of Gram matrix:
2060501320695  -233624820247
-296766663557  33648150444
  for  gp Pari: rt =  [ 2060501320695 , -296766663557 ; -233624820247 , 33648150444 ] ;    h =  [ 94 , 321 ; 321 , -74 ] ;    r =  [ 2060501320695 , -233624820247 ; -296766663557 , 33648150444 ] ;
=========================================

PARI/GP is free software, covered by the GNU General Public License,
?
? rt
%7 =
[2060501320695 -296766663557]
[-233624820247   33648150444]

? h
%8 =
[ 94 321]
[321 -74]

? r
%9 =
[2060501320695 -233624820247]
[-296766663557   33648150444]

? rt * h * r
%10 =
[ 94 321]
[321 -74]
</code>```

Title: Computing continued fraction expansions

Body:
took a while, I wrote the Gauss-Lagrange method using the left neighbor instead of the right neighbor. Given a reduced form $Ax^2 + B xy + C y^2,$  which means that $D= B^2 - 4 AC$ is positive but not a square, $AC < 0 $  and $B > |A+C|,$ the numbers listed that I call "epsilon" (well, its absolute value) give the partial quotients for the continued fraction of
$$ \frac{B+ \sqrt D}{2A}  $$
when $A$ is positive. I have not tried negative $A$ yet.. sample run

Sample: $47x^2 + 321 xy - 37 y^2$
and $D=109997,$  target is $ \frac{321+ \sqrt{ 109997}}{94}  $

First, a few "partial quotients" in the continued fraction expansion

? x = 321 + sqrt(109997)
%3 = 652.6579563345345104401576022
? x /= 94
%4 = 6.943169748239728834469761725
?
? x -= floor(x) ; x = 1/x
%5 = 1.060254531982535997122061997
? x -= floor(x) ; x = 1/x
%6 = 16.59626200880353974842520003
? x -= floor(x) ; x = 1/x
%7 = 1.677115068938572016859051397
? x -= floor(x) ; x = 1/x
%8 = 1.476853855235527410365538515
? x -= floor(x) ; x = 1/x
%9 = 2.097078568246198860000621270
? x -= floor(x) ; x = 1/x
%10 = 10.30093477958926629742181316
? x -= floor(x) ; x = 1/x
%11 = 3.322979156363579950723385592

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycleLeft  47  321  -37

0  form   47 321 -37   epsilon  6
1  form   -271 243 47   Epsilon  -1
2  form   19 299 -271   Epsilon  16
3  form   -191 309 19   Epsilon  -1
4  form   137 73 -191   Epsilon  1
5  form   -127 201 137   Epsilon  -2
6  form   31 307 -127   Epsilon  10
7  form   -97 313 31   Epsilon  -3
8  form   97 269 -97   Epsilon  3
9  form   -31 313 97   Epsilon  -10
10  form   127 307 -31   Epsilon  2
11  form   -137 201 127   Epsilon  -1
12  form   191 73 -137   Epsilon  1
13  form   -19 309 191   Epsilon  -16
14  form   271 299 -19   Epsilon  1
15  form   -47 243 271   Epsilon  -6
16  form   37 321 -47   Epsilon  8
17  form   -247 271 37   Epsilon  -1
18  form   61 223 -247   Epsilon  4
19  form   -163 265 61   Epsilon  -1
20  form   163 61 -163   Epsilon  1
21  form   -61 265 163   Epsilon  -4
22  form   247 223 -61   Epsilon  1
23  form   -37 271 247   Epsilon  -8
24  form   47 321 -37
  form   47 x^2  + 321 x y  -37 y^2

minimum was   19rep   x = -7   y = 1 disc 109997 dSqrt 331
Automorph, written on right of Gram matrix:
2060501320695  -233624820247
-296766663557  33648150444
  for  gp Pari: rt =  [ 2060501320695 , -296766663557 ; -233624820247 , 33648150444 ] ;    h =  [ 94 , 321 ; 321 , -74 ] ;    r =  [ 2060501320695 , -233624820247 ; -296766663557 , 33648150444 ] ;
=========================================

PARI/GP is free software, covered by the GNU General Public License,
?
? rt
%7 =
[2060501320695 -296766663557]
[-233624820247   33648150444]

? h
%8 =
[ 94 321]
[321 -74]

? r
%9 =
[2060501320695 -233624820247]
[-296766663557   33648150444]

? rt * h * r
%10 =
[ 94 321]
[321 -74]


问:
Fourier transform of compact supported function is entire
Suppose $f\in L^1(\mathbb{R})$ has compact support, say $\operatorname{supp}f\subset[-r,r]$.
I want to show that its Fourier transform
$$ \hat{f}(z) = \int_{-a}^{a} f(t) e^{-2\pi itz}dt$$
for $z\in\mathbb{C}$ is entire, so $\hat{f}\in H(\mathbb{C})$. I think this follows from Morera's theorem, but I am not very familiar with contour integrals. Could someone help me out with the first steps?

答:
"No, but I actually would like to learn how one could apply Morera's theorem here"

Well, for that we have to prove two things:

(a) $\hat f$ is continuous.

(b) The integral of $\hat f$ over each triangle is zero.

So, let's start with (a). Since for $z$ close to $z_0$,
$$
|e^{2\pi itz}-e^{2\pi itz_0}|\le 4\sqrt 2 \pi ae^{2\pi a(|1+Im(z_0)|)}|z-z_0|
$$
(that's at least what I could achieve), we have
$$
|\hat f(z)-\hat f(z_0)| = \left|\int_{-a}^af(t)\left(e^{2\pi itz} - e^{2\pi itz_0}\right)\,dt\right|\,\le\,4\sqrt 2 \pi ae^{2\pi a(|1+Im(z_0)|)}\|f\|_{L^1}|z-z_0|.
$$
Hence, $\hat f$ is locally Lipschitz. Now, let $\Delta$ be any triangle. Then with Fubini we get
$$
\int_\Delta\hat f(z)\,dz = \int_\Delta \int_{-a}^af(t)e^{2\pi itz}\,dt\,dz = \int_{-a}^af(t)\left(\int_\Delta e^{2\pi itz}\,dz\right)\,dt = 0.
$$


答:
The crux of the matter is:  $m$ and $n$ are both even if and only if both $\min(m, n)$ and $\max(m,n)$ are both even.

(That's obvious, isn't it.  If they are both even then the min and the max are one or the other of them and will have to be even.  And if the min and max are both even, each one is either the min or the max and the other will be the other extreme and they will both be even.

===

Let $\{p_i\}$ be all the prime factors of either $a$ or $b$.

We can write $a = \prod p_i^{m_i}$.  If it turns out that $p_k\not \mid a$ but $p_k\mid b$ this will be okay if we assume we can let $m_k = 0$.  And we can write $b = \prod p_i ^{n_i}$ (same caveat.)

Now $G=\gcd(a,b) = \prod p_i^{\min(m_i,n_i)}$ and $L=\operatorname{lcm}=\prod p_i^{\max(m_i,n_1)}$.

Now any one of those is a perfect square if and only if all the powers are even (and $0$ is an even number and $1$ is a perfect square so any concerns we have about what if $p_k$ divides one but not the other are addresses).

So.....

If $a$ and $b$ are both perfect squares then $m_i, n_i$ will all be even so $\min(m_i, n_i)$ and $\max(m_i, n_i)$ will all be even.  So $G=\gcd(a,b)$ and $L =\operatorname{lcm}(a,b)$ will both be perfect squares.

And if  $G=\gcd(a,b)$ and $L =\operatorname{lcm}(a,b)$ are both perfect squares then $\min(m_i, n_i)$ and $\max(m_i,n_i)$ are all even.  SO all $m_i, n_i$ are even, so $a$ and $b$ are both perfect squares.

We can even argue further (although we don't have to-- we are totally done!) that if either $a$ or $b$ aren't both perfect squares there will be at least one $m_i$ or at least one $n_i$ that is odd and that is true if and only if at least one $\max(m_i,n_i)$ or one $\min(m_i,n_i)$ that is odd, so at least one of $G=\gcd(a,b)$ and $L =\operatorname{lcm}(a,b)$ are not a perfect square.

======

The crux of the matter is:  $m$ and $n$ are both even if and only if both $\min(m, n)$ and $\max(m,n)$ are both even.

Title: Proving a and b are perfect squares if and only if $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$ are perfect squares

Body:
Let $a$, $b$ be two positive integers. Prove that $a$ and $b$ are both perfect squares if and only if both $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$ are perfect squares.

I believe the proof is based in prime factorization, as if $a$ is a perfect square, then $a = (p_1^{a_1}\cdot\ldots\cdot p_n^{a_n})^2$, and the same goes for the gcd and lcm when they are perfect squares. I'm just a little lost on how to connect the dots and prove both directions.


问:
How is the area of a country calculated?
As countries' or states' borders are not straight lines but they are irregular in nature, I wonder how anyone can calculate the area of a country or state.

When do you think the area of a country or state was first calculated?  Was it before satellites provided us accurate picture of the earth?

Note: I am not asking about surface area of a country. They are assumed as flat while calculating the area.

答:
One easy way to do this would be a Monte Carlo method. Say you had a map of a country with map area $A_{map}$. Throw a large number $N$ of small seeds randomly on the map. By the number of seeds to fall within the country $n$, one can find the area of the country:
$$A_{country} = A_{map} \cdot \frac{n}{N}$$
This of course ignores maps with projections that are nor area preserving, but never the less.</p>


问:
coin toss, possibly binomial distribution
We are given two coins, the first shows head with probability $p∈(0,1)$ while the second with probability $q∈(0,1)$. We toss these coins simultaneously and, if both show head, we stop. If not, we toss them again.
What is the probability that we need to toss the two coins at most k times before obtaining two heads?

I am unsure here of what distribution to use to solve this, and the parameters involved. At first I was thinking binomial distribution because it seems to fit the requirements for it, but I'm not entirely sure. Any advice would be greatly appreciated.

答:
The distribution you are facing is a geometric

$$X\sim Geo(pq)$$

defined on $x=1,2,3...$

Thus

$$P(X\leq k)=1-(1-pq)^k$$


问:
Prove that the universal covering space of a Lie group is unique
This is a problem in Lee's book, introduction to smooth manifold. I am trying to show that the universal covering space of a connected Lie group is unique.

Suppose $G$ is a Lie group, and $\tilde{G}$ and $\hat{G}$ are universal covering groups of $G$. Therefore, the smooth covering maps $\pi:\tilde{G}\to G$ and $\hat{\pi}:\hat{G}\to G$ are Lie group homomorphisms.

I am trying to show there exists a Lie group isomorphism $\Psi:\tilde{G}\to\hat{G}$ such that $\hat{\pi}\circ\Psi=\pi$. The following is my attempt.

I know a universal covering of a manifold is unique. It means that there exists a diffeomorphism of $\Psi:\tilde{G}\to\hat{G}$ such that $\hat{\pi}\circ\Psi=\pi$. What is left is to show $\Psi$ is actually a homomorphism, which exactly means that
$$\Psi\circ\tilde{m}=\hat{m}\circ(\Psi\times\Psi)$$
where $\tilde{m}$ and $\hat{m}$ are the multiplication maps in $\tilde{G}$ and $\hat{G}$, respectively.

I can show that both $\Psi\circ\tilde{m}$ and $\hat{m}\circ(\Psi\times\Psi)$ are lifts of the same map $m\circ(\pi\times\pi)$, where $m$ is the multiplication map of $G$. Hence, it suffices to show that $\Psi\circ\tilde{m}$ and $\hat{m}\circ(\Psi\times\Psi)$ agree at some point. I want to show they agree on $(\tilde{e},\tilde{e})$, where $\tilde{e}$ is the unity in $\tilde{G}$.

To prove that, it suffices to show $\Psi(\tilde{e})=\hat{e}$. Now, the proof is almost done. However, I cannot show it is true.

I made two futile following attempt.

First, I notice that $\pi(\tilde{e})=\hat{\pi}(\hat{e})$. Therefore, $\pi$ has a unique lifting map $F:\tilde{G}\to\hat{G}$ such that $F(\tilde{e})=\hat{e}$. Also, $\Psi$ is a lift of $\pi$. However, I do not know how to show they are the same.

Second, I notice that $\hat{\pi}\circ\Psi(\tilde{e})=e$. Therefore, I choose a neighborhood $U$ of $e$ evenly covered by $\hat{\pi}$. Clearly, $\hat{e},\Psi(\tilde{e})\in\hat{\pi}^{-1}(U)$. However, I do not know how to prove they belong to the same component of $\hat{\pi}^{-1}(U)$.

I know I must be missing something obvious, since I think the proof is almost done. Could anyone point it out? Thank you in advance.

答: The problem is that the map $\Psi$ is not unique, but you can force it to have $\Psi(\tilde e)=\hat e$ in a unique way. The universal covering space of a space is indeed unique up to isomorphism, but not up to a unique isomorphism. For instance, the covering map $\mathbf R \to S^1$ has many automorphisms (which are in bijection with $\pi_1(S^1) = \mathbf Z$). You obtain unicity of the isomorphism only in the category of pointed spaces. Thus, in fact, given the pointed space $(G,e)$ and your two covering spaces $(\tilde G, \tilde e)$ and $(\hat G, \hat e)$, the universal property of the covering space implies that there is a unique map of pointed spaces $\Psi: (\tilde G, \tilde e) \to (\hat G, \hat e)$.


问:
What is the relationship between Linear Combination and Linear System of Equations
My guess is that it has either to do with finding the solution to a system of linear equations. Honestly this question is so broad i'm confused, that's why i'm asking for help.</p>

答:
If $A$ is an $n\times n$ matrix, and $b$ is an $n\times 1$ column-vector, then solving the system $$Ax=b$$ is equivalent to expressing $b$ as a linear combination of the columns of $A$.  For example, if $n=3$ and then solving the system  $$\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{bmatrix}{x_1\\x_2\\x_3}\end{bmatrix}=\begin{bmatrix}{b_1\\b_2\\b_3}\end{bmatrix}$$ is equivalent to finding $x_1,x_2,x_3$ such that $$x_1\begin{bmatrix}a_{11}\\a_{21}\\{a_31}\end{bmatrix}
+x_2\begin{bmatrix}a_{12}\\a_{22}\\{a_32}\end{bmatrix}
+x_3\begin{bmatrix}a_{13}\\a_{23}\\{a_33}\end{bmatrix}=
\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}$$

as you can easily verify.


问:
Find the intersection point
For a Parabola   AB and CD are any two parallel chords having slope 1 .
C1 is a circle passing through O A B and C2 is a circle passing through O C D where O is the origin . C1 and C2 intersect at ?

We need to find the points where C1 and C2 intersect. I tried assuming a circle passing through OAB and using parametric from to generate a fourth degree equation. I was unable to proceed further . Can anyone give a hint ? I am in high school

答:
If $A=(t^2,2t)$ then $B=((2-t)^2,2(2-t))$ (I'm setting $a=1$). A generic circle passing through $A$, $B$ and $O$ has equation $x^2+y^2+cx+dy=0$. Substituting there the coordinates of $A$ and $B$ we find two equations, which can be solved for $c$ and $d$:
$$
c = -8 + 2 t - t^2, \quad d = 2 t - t^2.
$$
The intersections of circle and parabola can be found solving the system
$$
\cases{
x^2+y^2+cx+dy=0 \cr
y^2=4x \cr
}
$$
We can eliminate $x$ and insert the expressions for $c$ and $d$ found above, to obtain $$
y^4+\left(8 t-4 t^2-16\right) y^2+\left(32 t-16 t^2\right) y=0.
$$
But we already know three solutions of this equation:
$y=0$, $y=2t$ and $y=4-2t$, hence we
can factor it, to get:
$$
y\ (y-2 t)\ (y - 4 + 2 t)\ (4 + y)=0.
$$
We then obtain a fourth solution: $y=-4$, which is independent of $t$. Hence all circles pass through the point $(4,-4)$ (or, restoring units: $(4a,-4a)$).

Title: Find the intersection point


问:
Find all functions satisfying $ f \left( m ^ 2 + f ( n ) \right) = f ( m ) ^ 2 + n $, for all $ m , n \in \mathbb N $
Find all functions $f:\mathbb{N}\to\mathbb{N}$ such that
$$f(m^{2}+f(n))=f(m)^{2}+n\text {, for all }m,n\in \mathbb N$$

I was initially unable to solve this problem, so I referred to a hint. The hint is as follows:

Let $f(1)=k$. From the given equation, we get $f(m^{2}+k)=1+f(m)^{2}$ and $1+k^{2}=f(1+k)$. Use this to prove that
$$f(f(m))^{2}-f(k)^{2}=m^{2}-1 \ \ \ \ \ \ (*)$$
Taking $m=2$, conclude that $f(k)=1 \ \ \ (*)$.
 Obtain $f(f(m))=m$ for all $m\in \mathbb{N}$. Use this to prove that $$f(n+1)=f(n)+k^{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$$
By induction $f(n+1)=f(1)+nk^{2}=k+nk^{2}$. Conclude that $k=1$ and hence $f(n)=n$ for all $n\in \mathbb{N}$

This solution is very unclear to me. I have marked all unclear steps with $(*)$. Please breakdown this solution and provide well explained arguments for the marked steps. Please try to use the same notation I have provided in the solution for ease of understanding. Thanks for all help :)

答: Don't know how to solve but I can provide reasoning for the second step which you are unclear on. Plugging in $m=2$, $f(f(2))^2-f(k)^2=3$. Say $a=f(f(2))$ and $b=f(k)$. Thus, $a^2-b^2=(a-b)(a+b)=3$. Since the domain is the natural numbers, the only solution is $a=2$ and $b=1$. Since, $b$ is defined as $f(k)$, $f(k)=1$.


问:
Is this Wick rotation correct? (self-intersecting closed contour)
I wonder if what is shown in figure 9.1 here is correct?

Doesn't the contour self-intersect, i.e. it's not a simple closed curve hence the Residue theorem shouldn't apply to this closed contour, right?

答:
If you are referring to the statement

Since the contour $C$ does not enclose singularities, the integral over $C$ must vanish

this is correct. The contour consists of two simple closed curves, neither of which surrounds any singularities of the function. Therefore, the integral over each part is $0$, hence the total integral is $0$.


问:
Number of symmetric square matrices with 0/1 such that all rows and all cols contain at least one 1
Symmetric meaning the main diagonal, i.e., matrix[i][j] = matrix[j][i].

Examples: identity matrix; matrix filled with 1s; matrix where the first row and first col are all 1s

I am stuck because I can't find a good way to take into account the symmetric condition. I tried to simplify the problem into just the lower-left half, but I don't think that's possible. I also tried to think of a recursive approach, which works for the easier version of the problem without the symmetric condition, but I can't get it to work here.

答:
One recursive formula is to add one row at a time.  So find the number $f(m,n)$ of all $m \times n$ matrices with entries in $\{0,1\}$ and $m \leq n$ such that the first $m$ columns form a symmetric matrix and each row contains at least one $1$.  Let $f(0,n)=1$ for convenience, then $f$ satisfies the following recursion: $$f(m+1,n)=2^{n-m} f(m,n) - f(m,n-1).$$

This follows from the fact that the first $m$ entries of row $m+1$ are fixed by symmetry.  This leaves $2^{n-m}$ options for the rest of the row if we allow a row of all zeroes.  A row of all zeroes means that column $m+1$ of the $m\times n$ matrix must be all zeroes. So this column does not contribute any $1$ so there are $f(m, n-1)$ of such matrices.  This is the correction term that is subtracted in the recursion above.

The sequence $f(n,n)$ starts with (counting from $n=0$) $$1, 1, 5, 45, 809, \ldots$$ and matches sequence A322661 of the OEIS.  Indeed, the formulation there in terms of number of graphs is equivalent with the definition here: How many graphs with numbered vertices exist in which self-edges (loops) are allowed and in which each vertex has at least one edge connected to it.


问:
The value of the second-order Eulerian polynomials at x = -1/2.
Recently, the second-order Eulerian polynomials
$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n $
have been discussed on MSE [ a ,
 b ].

$$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n =
\sum_{k=0}^n \left\langle\!\!\left\langle n\atop k \right\rangle\!\!\right\rangle \, x^k $$

Here $\left\langle\!\!\left\langle n\atop k \right\rangle\!\!\right\rangle $
are the second-order Eulerian numbers A340556.
The values of these polynomials at $ x = \frac12 $ generate a sequence
that represents the solution of Schröder's fourth problem
(see MSE and
A000311).

That's a pretty nice result that suggests the question:
Do the values of the polynomials at $ x = -\frac12$ (times $ 2^n $) also have
a combinatorial meaning? That this might indeed be the case is a conjecture
suggested by many examples of combinatorial polynomials. The reader can get
an impression for himself while browsing through the OEIS.

Polynomials
P(x)
2^nP(1/2)
2^nP(-1/2)

Laguerre
A021009
A103194
A000262

Motzkin
A064189
A330796
A000244

BigSchröder
A080247
A065096
A239204

StirlingSet
A048993
A005493
A000110

StirlingCycle
A132393
A000254
A000774

Eulerian 1st
A173018
A180119
A001710

However, this question may not be easy to answer, so we ask a more concrete question.
The sequence $ 2^n \left\langle \!\left\langle - 1/2 \right\rangle \!\right\rangle_n $
can also be generated without reference to the second-order Eulerian polynomials by
series reversion.

$$ 2^n \left\langle\!\!\left\langle - \frac{1}{2} \right\rangle\!\!\right\rangle_n
= (n + 1)!\, [x^{n+1}]\, \text{Reversion}\left(\frac{6x + \exp(3x) - 1}{9}\right) \quad (n \ge 0). $$

Can someone confirm this equation?

Addendum:

The general form of the reversion is described by:
$$ \left\langle\! \left\langle x \right\rangle\! \right\rangle_n =
(n+1)!\, (1-t)^{2n + 1}\, [x^{n+1}]\,
\operatorname{Reversion}_{x}(x + t - t \exp(x)) \quad (n \ge 0) $$

答:
Here are some comments. Seeking to invert

$$-\frac{1}{9} + \frac{2}{3} x + \frac{1}{9} \exp(3x) = z$$

we consult Wikipedia on
LambertW
to find  that the closed form solution to

$$x = a + b \exp(cx)$$

is given by

$$x = a - \frac{1}{c} W(-bc \exp(ac)).$$

We thus write

$$x = \frac{3}{2} z + \frac{1}{6} - \frac{1}{6} \exp(3x).$$

to obtain

$$\frac{1}{6} + \frac{3}{2} z
- \frac{1}{3} W((1/2)\times \exp((9/2)z+1/2))
\\ = \frac{1}{6} + \frac{3}{2} z
- \frac{1}{3} W(\exp((1+9z)/2)/2).$$

We have

$$W(z) = \sum_{m\ge 1} (-1)^{m-1} m^{m-1} \frac{z^m}{m!}$$

We then obtain for $n\ge 1$

$$-\frac{1}{3} (n+1)! [x^{n+1}]
\sum_{m\ge 1} (-1)^{m-1} \frac{m^{m-1}}{m!}
\frac{\exp(m/2)}{2^m} \exp(9mx/2)
\\ = \frac{1}{3}
\sum_{m\ge 1} (-1)^{m} \frac{m^{m+n}}{m!}
\frac{\exp(m/2)}{2^{m+n+1}} 9^{n+1}
\\ = \frac{3^{2n+1}}{2^{n+1}}
\sum_{m\ge 1} (-1)^{m} \frac{m^{m+n}}{m!}
\frac{\exp(m/2)}{2^{m}}
.$$

Using the notation from the cited post we get for our closed form

$$\frac{3^{2n+1}}{2^{n+1}} Q_n(-\exp(1/2)/2).$$

Now with $T(-\exp(1/2)/2) = -1/2$ we get with the cited identity

$$\frac{3^{2n+1}}{2^{n+1}}
\frac{1}{(3/2)^{2n+1}}
\sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle
\left(-\frac{1}{2}\right)^{k}
\\ = 2^n \sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle
\left(-\frac{1}{2}\right)^{k}.$$

This is the claim. Note that with $n=0$ we get $Q_0(-\exp(1/2)/2) - 1$
from the series for a total of $\frac{3}{2} +
\frac{3}{2} (Q_0(-\exp(1/2)/2)-1) = \frac{3}{2} +
\frac{3}{2} \frac{1}{3/2} - \frac{3}{2} = 1$ which also agrees with
the polynomial.

Concerning the addendum. Inverting $x+t-t\exp(x)$ with respect to
$x$ we obtain

$$-W(- t \exp ( -t + z )) -t + z.$$

We then get for the proposed closed form with $n\ge 1$

$$- (1-t)^{2n+1} (n+1)! [x^{n+1}]
\sum_{m\ge 1} (-1)^{m-1} \frac{m^{m-1}}{m!}
(-1)^m t^m \exp(-tm) \exp(mx)
\\ = (1-t)^{2n+1}
\sum_{m\ge 1} \frac{m^{m+n}}{m!}
t^m \exp(-tm).$$

This is $(1-t)^{2n+1} Q_n(t\exp(-t)).$ Now we have $T(t\exp(-t)) = t$
so we obtain

$$(1-t)^{2n+1} \frac{1}{(1-t)^{2n+1}}
\sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle t^k
\\ = \sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle t^k$$

as claimed. We get for $n=0$, that the coefficient on the singleton $z$
is $(1-t)$ for a total of $1-t+ (1-t) (Q_0(t\exp(-t))-1) = 1-t + (1-t)
t/(1-t) = 1$ which is the correct value.


问:
Non $\sigma$-finite measure where $\int f g = 0$ for all $g\in \mathrm L^q$ but $f$ is not a.e. zero
On a measured space $(X,\mathcal A,\mu$) with $\mu$ being $\sigma$-finite one can show that if $\int_X fg \ \mathrm d \mu = 0$ for all $g\in \mathrm L^p$, then $f = 0$ $\mu$-a.e.

In an answer here, Danny Pak-Keung Chang provides an example where this is false, but the example is trivial in the sense that here $\mathrm L^p = \{0\}$. Could we come up with a non-trivial example using a non $\sigma$-finite measure ? I've tried for instance using a Hausdorff measure, say $\mathcal H^1$ over $\mathbb R^2$, but could not come up with anything.

答:
Suppose $f$ is a real-valued measurable function  such that $\int_X fg\; d\mu = 0$ for all $g \in {\rm L}^p$, but $f$ is not $\mu$-a.e. $0$.  There must be $\epsilon > 0$ such that
$\mu(f^{-1}((-\infty,-\epsilon) \cup (\epsilon,\infty))) > 0$.  WLOG we can assume
$\mu(f^{-1}((\epsilon,\infty)) > 0$.  Now if $f^{-1}((\epsilon, \infty))$ has a measurable subset $E$ with $0 < \mu(E) < \infty$, we could take $g$ to be the indicator function $I_E$ of $E$ and get $ \int_X f g \; d\mu = \int_E f \; d\mu \ge \epsilon \mu(E) > 0$ while $g \in L^p(\mu)$.  So the only possible counterexamples are the "trivial" ones where
there is a set of $\mu$-measure $+\infty$ that has no measurable subset
with finite and strictly positive measure.


问:
A confusion about iterated limits - What would be a rigourous definition?
The Dirichlet function is the indicator function $1_ℚ$ of the set of rational numbers $\mathbb{Q}$, i.e. $1_\mathbb{Q}(x) = 1$ if $x$ is a rational number and $1_\mathbb{Q}(x) = 0$ if $x$ is not a rational number (i.e. an irrational number).
What the professor asks to show is that :

$$ \forall x \in \mathbb{R},   1_\mathbb{Q}(x)= \lim_{n \rightarrow +\infty}(\lim_{m \rightarrow +\infty} |\cos(n!\pi x )|^m) $$

It seemed at first clear why the previous equality is true. But then it  raised this question in my head:

About iterated limits, taking a simple example, is it true to say the following?
$$\lim_{n \rightarrow +\infty}(\lim_{m \rightarrow +\infty} \frac{n}{m}) = 0  \ \ \text{but} \ \ \lim_{m \rightarrow +\infty}(\lim_{n \rightarrow +\infty} \frac{n}{m}) = +\infty$$
Another example:

$$\lim_{n \rightarrow +\infty}(\lim_{m \rightarrow +\infty} (1 +\frac{1}{n})^m) = +\infty  \ \ \text{and} \ \ \lim_{m \rightarrow +\infty}(\lim_{n \rightarrow +\infty} (1 +\frac{1}{n})^m) = 1 \\ \text{whereas} \ \ \lim_{n \rightarrow +\infty} (1 +\frac{1}{n})^n = e  $$

I hope I have been clear in my presentation.

Thanks in advance

答:
Yes - in general, the order in which we take limits matters. There are situations where limits commute (= order doesn't matter) but this should never be taken for granted.

Division is a bit of an odd case because we have to worry about dividing by zero; the issue already crops up with subtraction. We have $$\lim_{x\rightarrow+\infty}\lim_{y\rightarrow+\infty}x-y=-\infty\quad\mbox{but}\quad \lim_{y\rightarrow+\infty}\lim_{x\rightarrow+\infty}x-y=+\infty.$$

Both of your examples are right, but I think this is an easier one to start off with.


问:
Proof by contradiction and invalid reasoning
I am very aware of the several posts that are in this forum, but to be honest I am not quite satisfied with the provided answers. Also, I can tell that the best post is the following:

Proof by Contradiction, Circular Reasoning?

I would love to continue the discussion in that post, however my reputation level does not allow me to add any comments, and I still have some questions. Therefore, I want to continue discussing more about proof by contradiction and circular reasoning.

Suppose we wish to prove $(P \wedge Q) \rightarrow R$. We assume P and Q. Proof by contradiction begins by assuming $\neg R$, and from these two assumptions, a "contradiction" is derived, and sometimes that contradiction is that $\neg P$ or $\neg Q$ (our initial valid assumptions). Therefore, since there is a contradiction with our assumptions, we conclude that it cannot happen that $\neg R$, so it must be the case that $R$.

This proof strategy bothers me, because I have the feeling that there is some kind of circular (or invalid) reasoning going on there i.e. $(P \wedge Q) \rightarrow R$ is equivalent to $\neg R \rightarrow \neg(P \wedge Q)$, so if we already know that the consequent is false, in order for the statement to be true, $\neg R$ has to be false, so R is true, but R is part of the consequent part of the conditional statement (isn't doing this like assuming there is a biconditional i.e. because R is true, then $P \wedge Q$ is true?). I probably don't know how to explain it better, but how/why can we say that this is indeed a proof? Hence, my main question is why it is valid to assume $\neg R$ if our conclusion involves R? I have one explanation that kind of makes sense to me (still a little iffy though):

In more simple terms, where we want to prove that $P \rightarrow Q$, we are trying to do the following: $[(P \wedge \neg Q) \rightarrow Contradiction] \rightarrow (P \rightarrow Q)$ if we can prove that $(P \wedge \neg Q)$ leads to a Contradiction, then we are indeed proving that $P \rightarrow Q$, which was our original goal. Given that this is a tautology, whenever we follow this proof strategy structure, we are always proving $P \rightarrow Q$. However, saying that this is a tautology kind of bothers me for some reason.

I would sincerely appreciate your insights, and if there is anything you would improve in my post, please feel free to do so. I'm starting my formal journey in the proof realm (currently reading How To Prove It), and I have always heard that proofs by contradiction were the easiest strategy. However, I have learned that this strategy seems to be difficult at the beginning for all those who are trying to learn proof strategies.

答:
Proof by contradiction is really casework.

Case 1: The formal system is inconsistent. In this case, every statement is true (and also false, but we don't care about that). In particular, the statement we want to prove is true. Done!

Case 2: The formal system is consistent. So if we have a statement A that implies a contradiction, then statement A must be false, otherwise we would be in Case 1, but we are in Case 2. In your example, this means that "(P and Q) and (not R)" must be false, because it implies a contradiction. But "(P and Q) implies R" is literally defined as "[(P and Q) and (not R)] is false". So we just showed "(P and Q) implies R", which is what we wanted. Done!


问:
Why are the solutions for $\frac{4}{x(4-x)} \ge 1\;$ and $\;4 \ge x(4-x)\;$ different?
Why is $\;\dfrac{4}{x(4-x)} \geq 1\;$ is not equal to $\;4 \geq x(4-x)\;$?

Probably a really dumb question but I just don't see it :(

答:
Actually, $\frac4{x(4-x)}\geqslant1$ is equivalent to $4\geqslant x(4-x)$ if $x\in(0,4)$ since, in that case, $x(4-x)>0$.

But if $x>4$ or $x<0$, then $x(4-x)<0$, and therefore $\frac4{x(4-x)}\geqslant1\iff4\color{red}\leqslant x(4-x)$.


问:
Isolatedness of Lefschetz map - almost there
$f$ is a Lefschetz map on a compact manifold X. And I need to show the Lefschetz fixed point is isolated.

I proved that the graph of f is transversal to the diagonal inside $X \times X$, then I don't know how to proceed from here. Thank you very much for your help.

答:
Suppose $x_0$ is a Lefschetz fixed point. Take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood (after composing with proper coordinate functions) I can think of $f$ as a map from open ball in $\mathbb{R}^n,\,\,$ (say $B$) to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$  such that $f(0)=0.$ Consider the function $g=f-id.$ Then $g(0)=0$ and by lefschetz condition $det(dg)(0)\neq 0$ (as 1 is not an eigenvalue of $df$). Hence $g$ is a local diffeomorphism by inverse function theorem and we are done.<br>
Just out of curiosity are you reading the book differential topology by Guillemin and Pollack.  </p>

Body:
Suppose $x_0$ is a Lefschetz fixed point. Take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood (after composing with proper coordinate functions) I can think of $f$ as a map from open ball in $\mathbb{R}^n,\,\,$ (say $B$) to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$  such that $f(0)=0.$ Consider the function $g=f-id.$ Then $g(0)=0$ and by lefschetz condition $det(dg)(0)\neq 0$ (as 1 is not an eigenvalue of $df$). Hence $g$ is a local diffeomorphism by inverse function theorem and we are done.

Just out of curiosity are you reading the book differential topology by Guillemin and Pollack.


答:
Note that when you multiply by a negative number, you need to "switch" the equal sign. A very basic example:
$ -1 < 4 $ multiplied by $-1$ becomes $ 1 > -4$.

In your case you multiply both sides with $x(4-x)$ or $4x - x^2$. This is positive for $x\in(0,4)$. So this becomes:

$$ 4\leq x(4−x)\,for\,x\in[-\infty, 0] \cup [4,\infty] $$
$$ 4≥ x(4−x)\,for\,x\in[0, 4] $$

Title: Why are the solutions for $\frac{4}{x(4-x)} \ge 1\;$ and $\;4 \ge x(4-x)\;$ different?

Body:
Why is $\;\dfrac{4}{x(4-x)} \geq 1\;$ is not equal to $\;4 \geq x(4-x)\;$?

Probably a really dumb question but I just don't see it :(


答:
If $\dfrac4{x(4-x)}\ge 1$, $\:x(4-x)$ is positive, hence we can multiply both sides by $x(4-x)$ and therefore it implies $4\ge x(4-x)$.

However, the other way, $ 4\ge x(4-x)$ necessarily happens in particular if  $(4-x)<0$, and in this case it is equivalent to $\dfrac4{x(4-x)}\le 1$.


答:
Suppose, for example, $x < 0$, I.e. $x$ is negative.

This would mean that $4-x>4$ and therefore $x(4-x)<0$. In other words $x(4-x)$ is negative and therefore less than or equal to $4$.

Now, you know that:

$$4 \geq x(4-x)$$

As $x(4-x)$ is negative, you also know  $\frac{1}{x(4-x)}$ is negative so if you multiply the inequality above by $\frac{1}{x(4-x)}$ you have to flip the sign (as multiplying both sides of an inequality by a negative number means you have to flip the sign).

So you get:

$$\frac{4}{x(4-x)}\leq 1$$

Long story short: if you multiply an inequality by a negative number on both sides you must flip the sign around. This means when $x(4-x)<0$ the two inequalities aren't the same.


问:
I have n pairs of socks, each pair of a different color. In how many ways can I pair them?
A) the socks are asymmetrical (I must pair a left one with a right one)

This looks trivial:

I place the left socks in a random order. the order is immaterial, because I only care which 2 colors I get to wear.

then I place a right sock near each left sock

I can do this in $P(n) = n!$ ways

B) the socks are symmetrical (I can pair any sock with any other).

As I can wear any pair socks in both ways, I consider $(yellow,red)$ and $(red,yellow)$ as the same pairing.

I found several wrong answers from reasoning (fractional results, wrong result for trivial cases).

Ultimately, I am led to think that the two cases are actually the same problem, but can't find an explanation for that.

example of a wrong solution:

I choose n socks - can do this in ${2n \choose n}$ ways, and place them in a random order. the order is immaterial, because I only care which 2 colors I get to wear.

I can place the other n socks in $P(n) = n!$ ways

If I swap the two socks with the same colors, I have the same pairing. reduce by $\frac{1}{2^n}$

If I swap the two socks of a pair, I have the same pairing. reduce by $\frac{1}{2^n}$

result: $$ {2n \choose n}*n!*\frac{1}{2^{2n}} = \frac{ (2n)! }{n! 2^{2n}}$$

for $n=2$, this gives $3$, while the intuitive answer is $2$.

答:
There are $\frac{(2n)!}{2^nn!}$ ways to partition a set of $2n$ items into $n$ pairs. E.g., for $n=2$ there are $\frac{4!}{2^2\cdot2}=3$ ways to partition $4$ items into $2$ pairs. However, this calculation is based on the assumption that all $2n$ items are distinguishable, which is not the case in your second scenario: in that scenario the two socks of a given color are indistinguishable, so if red and yellow are the two colors, there is only one to make two mixed pairs, not two.

Your $\frac{(2n)!}{n!2^{2n}}$ does not actually give you $3$ for $n=2$: you have an extra factor of $2^n$ in the denominator, so it gives you $\frac{24}{32}=\frac34$.

Say that a pair of socks of the same color is a matched pair. One way to approach the problem is to count the pairings according to the number of matched pairs. For $k=0,\ldots,n$ there are $\binom{n}k$ ways to choose $k$ colors for $k$ matched pairs. The remaining $2n-2k$ socks of the other $n-k$ colors must be paired up so that no sock is matched with its mate. This amounts to choosing a derangement of these $n-k$ colors, something that can be done in

$$d_{n-k}=(n-k)!\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$

ways. Thus, the number of possible pairings is

$$\begin{align*}
\sum_{k=0}^n\binom{n}kd_{n-k}&=\sum_{k=0}^n\binom{n}k(n-k)!\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}\\
&=\sum_{k=0}^n\frac{n!}{k!}\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}\,.
\end{align*}$$

For $n=2$ this yields

$$\frac21\left(1-1+\frac12\right)+\frac21(1-1)+\frac22(1)=2\,,$$

just as it should.


答:
Initially I thought you meant how many different pairs could be worn and posted this.

A) It's n for the left foot times n for the right foot $n^2$

B) Imagine there are 4 colours.  Red Blue Green Yellow.

We can do RR RB RG RY (4), then we start with B, but have to leave off BR, already used.  So its BB BG BY (3), then start with G leaving off two used, its GG GY (2), then only YY (1).

They are the triangle numbers $0.5n(n+1)$

For all socks being paired answer B) still seems to involve the triangle numbers and becomes this...

to be continued

Title: I have n pairs of socks, each pair of a different color. In how many ways can I pair them?

Body:
A) the socks are asymmetrical (I must pair a left one with a right one)

This looks trivial:

I place the left socks in a random order. the order is immaterial, because I only care which 2 colors I get to wear.

then I place a right sock near each left sock

I can do this in $P(n) = n!$ ways

B) the socks are symmetrical (I can pair any sock with any other).

As I can wear any pair socks in both ways, I consider $(yellow,red)$ and $(red,yellow)$ as the same pairing.

I found several wrong answers from reasoning (fractional results, wrong result for trivial cases).

Ultimately, I am led to think that the two cases are actually the same problem, but can't find an explanation for that.

example of a wrong solution:

I choose n socks - can do this in ${2n \choose n}$ ways, and place them in a random order. the order is immaterial, because I only care which 2 colors I get to wear.

I can place the other n socks in $P(n) = n!$ ways

If I swap the two socks with the same colors, I have the same pairing. reduce by $\frac{1}{2^n}$

If I swap the two socks of a pair, I have the same pairing. reduce by $\frac{1}{2^n}$

result: $$ {2n \choose n}*n!*\frac{1}{2^{2n}} = \frac{ (2n)! }{n! 2^{2n}}$$

for $n=2$, this gives $3$, while the intuitive answer is $2$.


问:
Problem finding this equality
Let $X$ be a metric space, $Y \subset X$ a subset, and $x \in X$ a point.
We define the distance of $x$ from $Y$ as
$$d(x, Y) = \inf d(x,y)$$
Proof that
$$\bar{Y} = \{x \in X : d(x, Y) = 0\}$$

I know that
$$\bar{Y}=Y \cup D(Y) $$
Where $D(Y)$ are $Y$ Accumulation Points. Obviously if $y \in Y$ you have $d(y,Y)=0$. My doubt is how to try that $d(y,Y)=0$ if $y \in D(Y)$. Having tried this I can say that the demonstration is over?

答:
If $x$ obeys $d(x,Y)=0$, then either $x \in Y$ and then trivially $x \in \overline{Y}$, or $x \notin Y$. So assume we're in the latter case, and let $r>0$. Then $r$ is not a lower bound for $\{d(x,y): y \in Y\}$, or else $d(x,Y) = \inf  \{d(x,y): y \in Y\} \ge r >0$ (as inf is the largest lower bound) and we have $d(x,Y)=0$. Hence there is some $y' \in Y$ so that $d(x,y') < r$ to witness that $r$ is not such a lower bound, but then $B(x,r) \cap Y\setminus \{x\}$, as witnessed by this $y'$, so as $r>0$ was arbitrary, $X \in D(Y)$ and also $x \in \overline{Y}$. This shows one inclusion.

If $x \in \overline{Y}$, then $d(x,Y)=0$: $0$ is surely a lower bound for $\{d(x,y): y \in Y\}$ and suppose $r>0$ were a larger one. Then $B(x,r) \cap Y = \emptyset$ (or a point in the intersection would contradict immediately that $r$ were a lower bound for $\{d(x,y): y \in Y\}$). It follows that $x \notin Y$ and $x \notin D(Y)$ so $x \notin \overline{Y}$, contradiction. It follows that there can be no upperbound of $\{d(x,y): y \in Y\}$ that is $>0$ so $\inf \{d(x,y): y \in Y\} = d(x,Y)=0$ as required. This shows the other inclusion. (We could also have used the two cases for $x$ as well: $X \in Y$ then $d(x,x)=0 \in \{d(x,y): y \in Y\}$ so $d(x,Y)=0$ is a minimum; or $y \in D(Y)$ and for $r>0$ we pick

$y' \in B(x,r) \setminus \{x\}$ to show that $\inf \{d(x,y): y \in Y\} \le d(x,y') < r$ and as $r$ is arbitrary, $d(x,Y)=0$ that way too).

Title: Problem finding this equality


问:
Lower bound of $\sum_{i=0}^{\infty}\ln (1-\delta_i)$
I am reading this paper and in some point it is stated that

$$\sum_{i=0}^{\infty} \ln(1-\delta_i)\geq -\sum_{i=0}^{\infty} \frac{\delta_i}{1-\delta_i}\geq -\frac{1}{1-\delta_0}\sum_{i=0}^{\infty}\delta_i \geq -1 $$

with $\delta_i = \frac{L||\nabla f(x_k)||}{\lambda_n^2(f''(x_k))}$.

Could you please help to understand how this inequality is lower bounded by -1? Generally, any help to understand the general behavior of $\sum_{i=0}^{\infty} \ln(1-\delta_i)$ is highly appreciated.

答:
Note: In the paper you linked, the authors first prove that $\delta_0 \leq \frac{1}{4}$ and that $\delta_{k} \leq \frac{2}{3} \delta_{k-1}$, from which it is immediate that $\frac{\delta_i}{1 - \delta_i} \leq \frac{\delta_i}{1 - \delta_0}$ and
that all $\delta_i < 1$.

Since all $\delta_i < 1$, you can use the following inequality for the log:

$$
\log(1 + x) \geq \frac{x}{1 + x}, \quad \forall x > -1 \Rightarrow
\log(1 - \delta_i) \geq \frac{-\delta_i}{1 - \delta_i}
\geq -\frac{\delta_i}{1 - \delta_0}, \forall i.
$$

The last step is due to the fact that $\delta_i \leq \left(\frac{2}{3}\right)^i \delta_0$. In particular,

$$
\sum_{i=0}^{\infty} \delta_i \leq \delta_0 \sum_{i=0}^{\infty} \left( \frac{2}{3} \right)^i = \frac{\delta_0}{1 - \frac{2}{3}} = 3 \delta_0.
$$

Therefore,

$$-\frac{1}{1 - \delta_0} \sum_{i=0}^{\infty} \delta_i = - \frac{3\delta_0}{1 - \delta_0} \geq -1,
$$

after taking into account the range of $\delta_0 \leq \frac{1}{4}$.


问:
Faithful or free action in the definition of affine space
For a vector space $T$ over a field $K$, Bourbaki defines affine space as a homogeneous $T$-set $E$ on which $T$ acts faithfully. The definitions found both on Wikipedia and nlab say that the action should be free. Note that Bourbaki defines a faithful action as one which embeds $T$ into the symmetric group of $E$. Is this a typo? Does he mean free instead?

Edit: Even the french version of the book uses "fidèlement" instead of "librement".

答: If $T$ is an abelian group and $E$ is a homogeneous $T$-set, then $E$ is faithful iff it is free, so the two definitions are equivalent.  Indeed, the kernel of the action is the intersection of the stabilizers of all the points.  Since $E$ is homogeneous, the stabilizers of any two points are conjugate, which means they are equal since $T$ is abelian.  So the action is faithful iff the stabilizers are trivial, i.e. iff the action is free.


答:
Other answers (and comments) have already explained the potential pitfalls of multiplying both sides of an inequality by a quantity whose sign is sometimes negative. Here is an alternative approach that might help:

$$\begin{align}
{4\over x(4-x)}\ge1
&\iff{4\over x(4-x)}-1\ge0\\
&\iff{4-x(4-x)\over x(4-x)}\ge0\\
&\iff{4-4x+x^2\over x(4-x)}\ge0\\
&\iff{(2-x)^2\over x(4-x)}\ge0\\
&\iff x(4-x)\gt0\quad\text{(since }(2-x)^2\text{ is always non-negative)}\\
&\iff0\lt x\lt4
\end{align}$$

Remark: In general the quadratic in the numerator will not be a perfect square, in which case the final steps are a bit more complicated (you can wind up with more than one interval where the inequality holds). But the problem here was concocted to have a simple answer.


问:
doubt about basic definition
I have a very basic doubt regarding my last post understanding the basic definiton

Can anybody clear the point what would be the result if I take  $j=1$.

For ex. if I am taking $(a,x,1)$ and $(b,x,1)$ as the vertices, Will they be adjacent or not?
Here the value of $j$ is 1. I made them adjacent as acc to definition...

vertices $(x_1, x_2, . . . , x_k)$, $(y_1, y_2, . . . , y_k)$ are adjacent if for some index j $\in$  {1, 2, . . . , k} we
have $x_jy_j$ $\in$$E(G_j )$ and $x_i = y_i$ for each 1$\leq$ i < j.

In example I took $V(P_3)$ = {a,b,c,d}, $V(K_2)$= {x,y} and $V(K_2)$={1,2}.
And the product is $P_3$$\circ$$K_2$$\circ$$K_2$

Thanks a lot.

答:
If $x_1$ and $y_1$ are adjacent in $G_1$, then $\langle x_1,x_2,\dots,x_k\rangle$ and $\langle y_1,y_2,\dots,y_k\rangle$ are adjacent in $G_1\circ G_2\circ\ldots\circ G_k$ no matter what $x_2,\dots,x_k$ and $y_2,\dots,y_k$ are. In this case the condition that $x_i=y_i$ for $1<i\le j$ is vacuously satisfied, because there are no such $i$. Thus, in your example $\langle a,u,m\rangle$ is adjacent to $\langle b,v,n\rangle$ in $P_3\circ K_2\circ K_2$ for any choice of $u,v\in\{x,y\}$ and $m,n\in\{1,2\}$.

Here’s another way to describe the adjacency relation in $G_1\circ G_2\circ\ldots\circ G_k$. Given distinct vertices $x=\langle x_1,x_2,\dots,x_k\rangle$ and $y=\langle y_1,y_2,\dots,y_k\rangle$, let

$$d(x,y)=\min\Big\{i\in\{1,\dots,k\}:x_i\ne y_i\Big\}\;;$$

then $x$ and $y$ are adjacent in $G_1\circ G_2\circ\ldots\circ G_k$ if and only if $x_{d(x,y)}$ and $y_{d(x,y)}$ are adjacent in $G_{d(x,y)}$.

In words, examine the coordinates of $x$ and $y$ from left to right. Since $x\ne y$, you will eventually come to one at which $x$ and $y$ differ. Say that this first occurs at position $j$: then $x$ and $y$ are adjacent in $G_1\circ G_2\circ\ldots\circ G_k$ if and only if and only if $x_j$ and $y_j$ are adjacent in $G_j$. (This $j$ is the $d(x,y)$ of the previous paragraph.) This $j$ might be $1$, in which case you don’t even have to look at any other coordinate: if $x_1$ and $y_1$ are adjacent in $G_1$, then $x$ and $y$ are adjacent in $G_1\circ G_2\circ\ldots\circ G_k$. Or $x_1$ and $y_1$ might be equal, in which case you go on and look at $x_2$ and $y_2$. If $x_2\ne y_2$, then $x$ and $y$ are adjacent in $G_1\circ G_2\circ\ldots\circ G_k$ if and only if $x_2$ and $y_2$ are adjacent in $G_2$; if $x_2=y_2$, you go on and look at $x_3$ and $y_3$. And so on.

Title: doubt about basic definition


问:
About mollifiers
When we define mollifiers as in Evans's book, we use in one dimension say, $\phi_\epsilon=(1/\epsilon)\phi(x/\epsilon)$ where $\phi$ is standard mollifier. My question is what is the rate of convergence relative to $\epsilon$ for the norm $\|f_\epsilon-f\|_{L^1}$ where $f_\epsilon=\phi_\epsilon*f$?

答:
The answer is not universal, it depends on the properties of the integrable function $f$. Suppose for example that $f$ is Lipschitz continuous with constant $K$. Then

$$||f_\epsilon-f||_1=\int_{-\infty}^{\infty}\frac{1}{\epsilon}\phi\left(\frac{t-t'}{\epsilon}\right)|f(t')-f(t)|dt'\leq K\epsilon\int_{-\infty}^{\infty}x\phi(x)dx$$

and the rate of convergence is at least $\mathcal{O}(\epsilon)$. If you choose $f$ to be a generic integrable function, then you can show that, in one dimension at least, the rate of convergence may be non-uniform and in fact arbitrarily slow as $\epsilon\to 0$ for certain points. A counterexample is given by the function $f(x)=|x|^{a}, 0<a<1$, which is integrable in any arbitrary interval containing the origin. Choose $t=0$ and then the norm in question reads

$$||f_\epsilon-f||_1=\epsilon^a\int_{-\infty}^{\infty}|x|^{a}\phi(x)dx$$

Also, the convergence rate may be sped up arbitrarily, if several derivatives of $f$ are zero at a certain point. In fact you can show that $||f_{\epsilon}(t_0)-f(t_0)||_1=\mathcal{O}(\epsilon^n)$ where $n$ is the order of the first non-zero derivative at $t=t_0$.

Title: About mollifiers


问:
Why does the dicyclic group have exactly one involution?
This is Exercise 12.2(a) of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search, it is new to MSE.

The Details:

Roman defines, on page 350 ibid., the dicyclic group of order $4n, n>1$, using the presentation

$${\rm Dic}_n\cong \langle x,y\mid x^{2n}, y^2=x^n, yx=x^{2n-1}y=x^{-1}y\rangle.\tag{$\mathcal{P}$}$$

The Question:

Why does the dicyclic group have exactly one involution?

Thoughts:

Since the order of the dicyclic group is $4n$, two divides its order, so by Cauchy's Theorem, an involution exists; but this is nuking the problem, since $y^2=x^n$ is an involution by the relation $x^{2n}=1$ in $(\mathcal{P})$.

Suppose, then, for a contradiction, that $z^2=1$ for some $z\in{\rm Dic}_n$ with $z\neq x^n$. Now what do I do?

Experience tells me that we can write each element of ${\rm Dic}_n$ in the form $x^ay^b$ for some $a\in\overline{0,2n-1},$ $b\in\{0,1\}$. This follows from the orders of $x$ and of $y$, together with the last relation in $(\mathcal{P})$. I'm fairly sure this will help.

An Example:

Let $n=2$. Then ${\rm Dic}_2$ is isomorphic to the quaternion group. Clearly $-1$ is its only involution.

My Background:

I work with presentations a lot, so this seems like a problem I should be able to solve myself with a bit of a hint. I don't have much experience with dicyclic groups.

Further Context:

This question is from a rather introductory chapter titled "Free Groups and Presentations", so the tools available aren't as sophisticated as in, say, Magnus et al.'s "Combinatorial Group Theory: [. . .]".

Please help :)

答:
Well, it is clear that one involution is $x^n$, and this is the only power of $x$ that is an involution. Every other element can be written in the form $x^iy$ with $0\leq i\lt 2n$, using the given relations. Note that any even power of $y$ can be written as a power of $x$, since $(y^{2k}) = (y^2)^k = (x^n)^k = x^{kn}$, and any odd power of $y$ can be written as a power of $x$ times $y$, since $y^{2k+1} = y^{2k}y = x^{kn}y$.  That’s why every element can be written as $x^iy^j$ with $0\leq i\lt 2n$, $0\leq j\lt 2$. Uniqueness takes a bit more work, but isn’t terribly hard.

We have
$$(x^iy)^2 = x^iyx^iy = x^ix^{-i}y^2 = x^n.$$
So this is trivial if and only if $x^n=e$, which cannot happen since $x$ has order $2n$. Thus, none of these elements are involutions, and so the only involution is $x^n$, which we already found.


问:
Do closure and intersection of open sets commute when composed with the interior?
In the Wikipedia article on the interior operator it is stated that on a complete metric space $X$ one has for every sequence of open sets $(A_i)_{i\in \mathbb N}\subset X$ the relation
$$
\mathrm{Int}\Big(\bigcap_{i}\overline A_i\Big)=\mathrm{Int}\Big(\overline {\bigcap_{i}A_i}\Big).
$$
Now I wonder whether the finite version of this relation, i.e., for any two open sets $A,B\subset X$ to have
$$
\mathrm{Int}\big(\overline A\cap \overline B\big)=\mathrm{Int}\big(\overline {A\cap B}\big),
$$
does actually hold in any topological space $X$. If so, it should be found in some elementary textbooks, perhaps as an exercise. I searched for a while but it turned out that searching for relations like this is extremely inconvenient - one obtains hundreds of results which are similar but different from the wanted relation. On the other hand, I guess that (if the relation holds in general) there should be an easy direct proof based on playing around with the operations interior, closure, intersection, union, and complement. Since I don't have enough time for  this game I'm asking directly for a reference.

答:
Yes, it's a standard fact used in the theory of so-called regular-open sets (sets such that $\operatorname{int}(\overline{U})=U$; your sets of the form "interior of closure" are regular open..) and forms part of the proof that those sets form a Boolean algebra under the operations $U \land V = U \cap V$, $U \lor V = \operatorname{int}(\overline{U \cup V})$ and $\lnot U = \operatorname{int}U^\complement$. I think Halmos proves it in his Boolean algebra book, and it's also in Engelking somewhere, I'm sure.

As $\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$, because $A \subseteq \overline{A}$ and likewise for $B$, so $\overline{A} \cap \overline{B}$ is a closed superset of $A \cap B$, so it contains $\overline{A \cap B}$ too. Taking interior on both sides gives the easy inclusion (we don't need anything on $A$,$B$):

$$\operatorname{int}(\overline{A \cap B})\subseteq \operatorname{int}(\overline{A} \cap \overline{B})$$

For the reverse inclusion we can use a small lemma that for an open $U$ and any $A$ we have $\overline{U \cap A} = \overline{U \cap \overline{A}}$. Think about it.


问:
Relative Condition number of composite function
I want to find a function $$h = g \circ f$$ such that condition number of $g$ and $f$ are greater than $10$, but the condition number of $h$ is less than $1$.

I am trying to use polynomials like $x^{10}$ for $f$ and $g$, but I find that the condition number of $h$ can't be less than $1$.
If I fix the condition number of $h$ to be less than $1$ and the condition number of $f$ to be greater than $10$, then I find that the condition number of $g$ must be less than $1$.

答:
Typically, the condition number of a composition is the product of the condition numbers. Specifically, if $x \not =0$ and $h(x) \not = 0$, then the relative condition number of the function $h$ at the point $x$ is given by
$$ \kappa_h(x) = \left | \frac{xh'(x)}{h(x)} \right|.$$
Now if $h = f \circ g$ and the chain rule can be applied, then
$$ \frac{xh'(x)}{h(x)} = \frac{x f(g(x))g'(x)}{f(g(x))} = \frac{y f'(y)}{f(y)}\frac{x g'(x)}{g(x)}$$
where $y = g(x)$ and we assume that $y \not = 0$. It follows that
$$\kappa_h(x) = \kappa_f(g(x)) \kappa_g(x).$$
In particular, if $\kappa_f(g(x)) \ge 10$ and $\kappa_g(x) \ge 10$, then $\kappa_h(x) \ge 100$ is inevitable.


答:
The difference is:

We can have $4 \ge x(x-4)$ if $x(x-4) \le 0$.

But we can not have $\frac 4{x(x-4)} \ge 1$ if $x(x-4) \le 0$.

(if $x(x-4) = 0$ then $\frac 4{x(x-4)}$ doesn't exist.  And if $x(x-4) < 0$ then $\frac 4{x(x-4)} < 0 < 1$.)

(Otherwise, when $x(x-4) > 0$ then $\frac {4}{x(x-4)} \ge 1$ and $4 \ge x(x+4)$ are the same thing.)

So $4\ge x(x-4)$ is true more often and will have more solutions than $\frac 4{x(x-4)}\ge 1$.

(Although, come to think of it,  $4 \ge x(x-4) > 0$ and $\frac 4{x(x-4)} \ge 1$ will have the exact same solutions.....)

======

To solve $4 \ge x(x-4)$ we have

$x(x-4) = x^2 - 4x \le 4$ so

$x^2 - 4x + 4 \le 8$ so

$0 \le (x -2)^2 \le 8$ so

$0 \le x - 2 \le \sqrt 8$ so

$2-\sqrt 8 \le x \le \sqrt 8+2$

But notice these solutions will include cases where $x(x-4) \le 0$.

.....

TO solve $\frac 4{x(x-4)}\ge 1$ we must consider three cases:

Case 1:  $x(x-4) < 0$.

Then $\frac 4{x(x-4)} \ge 1 \implies$

$\frac 4{x(x-4)} x(x-4) \le 1\cdot x(x-4)\implies$

$4 \le x(x-4) < 0$.  But that is impossible.

Case 2: $x = 0$

Then $\frac 4{x(x-4)} = \frac 4 0$ and that is impossible.

Case 3: $x(x-4) > 0$.

Then we have either $x > 0$ and $x-4 > 0$; so $x> 4$.   Or $x < 0$ and $x-4 < 0$; so $x < 0$.  So we are restricted to requiring $x > 4$ or $x < $.  In other words $0 \le x \le 4$ is impossible.

But if we make that restriction then

$\frac 4{x(x-4)} \ge 1 \implies $

$\frac 4{x(x-4)} x(x-4) \ge 1\cdot x(x-4) \implies$

$4 \ge x(x-4)$.

But note !!!VERY IMPORTANT!!!! we can only do this because we are assuming $x(x-4) > 0$.  If we didn't know $x(x-4) > 0$ we could not say this.

Now we solved $4 \ge x(x-4)$ above and got $2-\sqrt 8 \le x \le 2 +\sqrt 8$.

But we MUST combine it with $x \not \in [0, 4]$.

So the solution here is $2-\sqrt 8 \le x < 0$ or $4 < x < 2 + \sqrt 8$.

.....

These two solutions are the same EXCEPT $4 \ge x(x-4)$ includes cases where $x(x-4) \le 0$ (which occures if and only if $0\le x \le 4$)  while  $\frac 4{x(x-4)} \ge 1$ does not include those.

[SO as the solutions to $4\ge x(x-4)$ is $[2-\sqrt 8, 2 + \sqrt 8]$ then the solutions to $\frac 4{x(x-4)} \ge 1$ is $[2-\sqrt 8, 2+\sqrt 8] \setminus [0, 4] = [2-\sqrt 8, 0) \cup (4, 2+\sqrt 8]$.


问:
demonstrate the following argument
argument to be demonstrated:

$A ∨ (B ∧ C)$

$B → D$

$C → E$

$D ∧ E → A ∨ C$

$¬A$

$∴ C$

My attempt at this mathematical demonstration was as follows:

$A ∨ (B ∧ C)$

$≡ (A ∨ B) ∧ (A ∨ C)$

$≡ (A ∨ A) ∧ (B ∨ C)$

$≡ A ∧ (B ∨ C)$

$≡ (A ∧ B) ∨ (A ∧ C)$

答:
This is a deductive reasoning question, rather than a propositional calculus one.

You are promised: $A\lor(B\land C)$, $B\to D$, $C\to E$, $(D\land E)\to (A\lor C)$, and $\lnot A$.

Well, because $A\lor (B\land C)$ and $\lnot A$, therefore $B\land C$, via Disjunctive Syllogism.

Because $B\land C$, therefore $B$ and $C$, via Simplification.

Because $B$ and $B\to D$ , therefore $D$, via Modus Ponens.

Bec...  and so on.

Title: demonstrate the following argument


问:
Artin-Schreier Question from Corps Locaux
I have a question from Serre's book "Corps Locaux", namely question 5a in section 2 of chapter IV. It is as follows:

"Let $e_K$ be the absolute ramification index of K, and let n be a positive integer prime to p and (strictly) less than $pe_K/(p-1)$; let y be an element of valuation -n. Show that the Artin-Schreier equation $x^p - x =y$ is irreducible over K, and defines an extension L/K which is cyclic of degree p. (Show that if x is a root of this equation, then the other roots have the form $x + z_i$ $(0 \leq i < p$, with $z_i \in A_K$, and $z_i \equiv i \mod{\mathbb{p_L}}$."

Here $K$ is characteristic 0, its residue field has positive characteristic p, and $K$ is a complete local field.

I'm struggling with the last part, namely showing the roots have the given form. Right now, as $(n,p)=1$, if $x$ is any root then $val(x^p) = -n$, which forces any root to not be within $K$. If I can show that in the extension of $K$ generated by $x$ I can write all other roots in the desired form, this will mean $K(x)$ is the splitting field and the Galois group will act transitively on the roots, whence the result.

Could I be pointed in the right direction?

Thanks,

Garnet

答:
We will show that the other roots are of the form $x + z_i$ for $z_i \in A_L$, the integral closure of $A_K$ in $A_L$. To do this, we just need to show the $z_i$ are roots of a monic equation with coefficients in $A_L$, since $A_L$ is integrally closed. Our only condition is $x^p - x - y = 0$, so we take the difference of $x^p-x-y$ and $(x+z)^p - (x+z) - y$ to obtain
$$ 0 = (x+z)^p - x^p - (x+z) + x = -z + \sum_{i=1}^p \left(\binom{p}{i} x^{p-i}\right) z^i.$$
Now use your conditions to show that this equation is what we desire.

References: "Local Fields and Their Extensions" by I.B. Fesenko and S.V. Vostokov, III.2.5


问:
Proof explanation for how this criteria shows exactness
This is about a proof from Homological Algebra on a Complete Intersection... by Eisenbud.

Here, $A$ is a Noetherian commutative ring.

To show exactness at $\bar F$, we must show that $\operatorname{Im} \bar \psi \subset \ker \bar \varphi$ and that $\ker \bar \varphi \subset \operatorname{Im} \bar \psi$.

How does the highlighted text below imply this?

答:
Since $\varphi \psi = x$, reducing modulo $x$ we find $\bar{\varphi} \bar{\psi} = 0$, so $\operatorname{im} \bar{\psi} \subseteq \ker \bar{\varphi}$.

Conversely, let $\bar{u} \in \ker \bar{\varphi}$, and $u$ any lift of $\bar{u}$ to $F$.  Take $\kappa: A \to F$ determined by $\kappa(1) = u$, so that $\bar{\kappa}(1) = \bar{u}$. Then by assumption $\bar{\varphi} \bar{\kappa} = 0$.  Suppose we could show from this that $\bar{\kappa}$ factors through $\bar{\psi}$, i.e., $\bar{\kappa} = \bar{\psi} \bar{\gamma}$ for some $\bar{\gamma}: B \to \bar{G}$.  Then $\bar{u} = \bar{\kappa}(1) = \bar{\psi} (\bar{\gamma}(1))$, so $\bar{u} \in \operatorname{im} \bar{\psi}$.  Therefore, $\ker \bar{\varphi} \subseteq \operatorname{im} \bar{\psi}$.


问:
Interesting logarithmic inequality
Recently the following question was published on the site and soon after this deleted:

Prove the inequality
$$
\left(\log\frac{a^2+b^2}{2ab}\right)^n\le \left(\log\frac{a^2+c^2}{2ac}\right)^n+\left(\log\frac{b^2+c^2}{2bc}\right)^n\tag1
$$
for all positive $a,b,c$ and $n=\frac12$.

The inequality has many nice features. Particularly $\frac12$ seems to be the largest possible value of $n$ such that (1) generally holds.

I have succeeded to prove the inequality in a rather awkward way but I believe there should be a simple and nice way to prove it. Possibly it is just a particular case of a more general inequality.

Any hint is appreciated.

Title: Interesting logarithmic inequality

Body:
My attempt :

$\sqrt{ln (\frac{1+(\frac{b}{a}) ^2}{2\frac{b} {a} })}<\sqrt{ln (\frac{1+(\frac{c}{a}) ^2}{2\frac{c} {a} })}+
\sqrt{ln (\frac{1+(\frac{c}{b}) ^2}{2\frac{c} {b} })} $
Suppose $x=\frac{b}{a} $ and $y=\frac{c} {a} $

$\sqrt{ln (\frac{1+x^2}{2x})}<\sqrt{ln (\frac{1+y^2}{2y})}+\sqrt{ln (\frac{x^2+y^2}{2xy})}$

$\Leftrightarrow$$ln (\frac{1+x^2}{2x})<ln (\frac{1+y^2}{2y})+ln (\frac{x^2+y^2}{2xy})+2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})} $

$\Leftrightarrow$$ln (\frac{1+x^2}{2x})-ln (\frac{1+y^2}{2y})-ln (\frac{x^2+y^2}{2xy})-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $
$\Leftrightarrow$$ln (\frac{2y(1+x^2)}{2x(1+y^2})-ln (\frac{x^2+y^2}{2xy})-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $
$\Leftrightarrow$$ln (\frac{2y(1+x^2)(2xy) }{2x(1+y^2)(x^2+y^2)}-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $
$\Leftrightarrow$$ln (\frac{2y^2(1+x^2) }{(1+y^2)(x^2+y^2)}-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $
$\Leftrightarrow$$ln (\frac{(1+x^2)2y^2 }{(1+y^2)(x^2+y^2)}-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $

So that is always correct if $ ln(\frac{(1+x^2)2y^2 }{(1+y^2)(x^2+y^2)}$<0
$\Rightarrow$$\frac{(1+x^2)2y^2 }{(1+y^2)(x^2+y^2)}<1$

$\Rightarrow$$y^2x^2+y^2-x^2-y^4<0$$\Rightarrow$$ (x^2-y^2)(y^2-1)<0$$\Rightarrow$$ x<y<1$ or $x>y>1$

that is correct if

$b<c<1$ or $b>c>1 $

So I am showed this inequality for $b<c<1 $ and $b>c>1 $


问:
Non-Homogenous 2nd ODE
Given the equation: $2y''+3y'+y=(x+2)e^{-x}$

I am wondering how you could go about dealing with this forcing term when they are multiplied.

I understand if it was: $(x+2) + e^{-x}$ but that isn't the case! Thanks.

Title: Non-Homogenous 2nd ODE

Body:
$$2y′′+3y′+y=(x+2)e^{−x}$$
$$2(y′′+y')+(y′+y)=(x+2)e^{−x}$$
$$2(e^xy′′+e^xy')+(e^xy′+ye^x)=(x+2)$$
Rewrite the DE as:
$$2(e^xy')'+(e^xy)'=(x+2)$$
$$(2e^xy'+e^xy)'=x+2$$
$$(2(e^xy)'-e^xy)'=x+2$$
$$(2u'-u)'=x+2$$
With $u=e^xy$. And integrate.


问:
Cumulative distribution function with 3 variables
Let $X$ be the random variable whose cumulative distribution function is
$$
F_X (x) = \begin{cases}
    0, & \text{for} \space x\lt 0 \\
    \frac{1}{2}, & \text{for} \space 0\le x\le 1 \\
    1, & \text{for} \space x\gt 1 \\
\end{cases}.
$$
Let $Y$ be a random variable independent of $X$ and uniformly distributed over the interval $(0,1)$. Define the random variable $Z$ as
$$
Z = \begin {cases}
    X, & \text{if} \space X\le \frac{1}{2} \\
    Y, & \text{if} \space X\gt \frac{1}{2} \\
\end{cases}
$$
Determine $\mathbb{P} (Z\le \frac{1}{5})$.

I believe that $X$ only takes the discrete values $0$ and $1$ with equal probability, but I'm not entirely sure. By intuition, I think that the answer is $\frac{1}{2}$. I'm unsure about this question, so any advice would be appreciated.

答:
You should define $F(1)$ as $1$ instead of $\frac 1  2$.

$$P(Z \leq \frac 1  5)$$ $$=P(X \leq \frac 1  5)+P(Y \leq \frac 1  5, X>\frac 1  2)$$ $$=P(X \leq \frac 1  5)+P(Y \leq \frac 1  5)P( X>\frac 1  2)$$

$$=\frac 1 2 +\frac 1 5  (1-\frac 1 2 )=\frac 3 5.$$


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

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

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

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

答: Almost everything I've seen so far, but especially that the area under Gaussian curve converges to the square root of the ratio of circumference of the circle to its diameter. This result is old and well-known, but these two things seem so unrelated that I still find it amazing! </p>


答:
I believe that $X$ only takes the discrete values $0$ and $1$ with equal probability, but I'm not entirely sure.

Yes, that is what the cumulative distribution function is saying: as it increases suddenly at $0$ and $1$, and remains constant elsewhere.  Thus $X$ has the probability mass function of:

$\qquad\mathsf P(X=x)=\begin{cases}1/2 &:& x=0~\text{or}~ x=1\\0&:&\text{elsewhere}\end{cases}$

So you should see that when $X=0$ then $Z=0$, or when $X=1$ then $Z=Y$, so: $$\mathsf P(Z\leq 1/5)=\mathsf P(X = 0)+\mathsf P(X=1\cap Y\leq 1/5)$$

Title: Cumulative distribution function with 3 variables

Body:
Let $X$ be the random variable whose cumulative distribution function is
$$
F_X (x) = \begin{cases}
    0, & \text{for} \space x\lt 0 \\
    \frac{1}{2}, & \text{for} \space 0\le x\le 1 \\
    1, & \text{for} \space x\gt 1 \\
\end{cases}.
$$
Let $Y$ be a random variable independent of $X$ and uniformly distributed over the interval $(0,1)$. Define the random variable $Z$ as
$$
Z = \begin {cases}
    X, & \text{if} \space X\le \frac{1}{2} \\
    Y, & \text{if} \space X\gt \frac{1}{2} \\
\end{cases}
$$
Determine $\mathbb{P} (Z\le \frac{1}{5})$.

I believe that $X$ only takes the discrete values $0$ and $1$ with equal probability, but I'm not entirely sure. By intuition, I think that the answer is $\frac{1}{2}$. I'm unsure about this question, so any advice would be appreciated.


问:
Probability problem about a card located in a wardrobe or in $1$ of $5$ drawers each of them have the same probability

A card is either in a wardrobe with a probability of $\frac{1}{5}$ or in one of five drawers with the same probability each drawer.
If we have opened four drawers, chosen at random, and the card is not there, what is the probability that it is in the fifth drawer?

答:
If $W$ is the probability of the card being in the wardrobe, and $D$ is the probability of it being in the remaining drawer, we have
$$P(D\mid W\cup D)=\frac{P(D\cap(D\cup W))}{P(D\cup W)}=\frac{P(D)}{P(D\cup W)}=\frac{\frac{4}{25}}{\frac{9}{25}}=\frac49$$
One notes that this is the ratio of the original probabilities of being in that particular drawer to being in that drawer or the wardrobe.


问:
Convolution is a Lipschitz Map
Consider a well-behaved space $S$ of probability distributions on $\mathbb{R}^d$, and consider the $p$th Wasserstein metric $d_p : S \times S \rightarrow \mathbb{R}_{\ge 0}$. For a fixed distribution $\nu \in S$ with a bounded support (which represents error), is the map $\mu \mapsto (\mu * \nu)$ Lipschitz with respect to the $p$th Wasserstein distance? For which $p$?

I suppose a part of the question is figuring out what such a well-behaved space $S$ could be, and I'd like to apologize ahead because I'm not an expert in analysis or probability theory.

答:
Actually, if we take $S$ to be the whole space of probability distributions on $\mathbb{R}^d$, even with no restrictions on $\nu$ (besides that $\nu$ is a probability distribution), the map $\mu \mapsto \mu * \nu$ is Lipschitz with Lipschitz constant at most 1.

Recall the definition of the $p$-th Wasserstein distance (on the space of probability measures):
$$ d_W(\mu, \mu') =
\inf \left(\mathbb{E}[ \| X - Y \|^p]\right)^{1/p}, $$
where the infimum is taken over all joint distributions on random variables $(X,Y)$ such that $X$ is $\mu$-distributed and $Y$ is $\nu$-distributed.

Recall also that if $X$ is $\mu$-distributed and $Z$ is $\nu$-distributed and $X$ and $Z$ are independent, then $X+Z$ is $\mu * \nu$ distributed.
Thus, whenever $X$ is $\mu$-distributed and $Y$ is $\mu'$-distributed, if we take $Z$ to be a $\nu$-distributed random variable which is independent of $(X,Y)$, then $(X+Z,Y+Z)$ has marginal distributions $\mu * \nu$ and $\mu' * \nu$ respectively. Thus for any such $(X,Y)$ we get the bound

$$ d_W(\mu * \nu, \mu' * \nu)^p \le \mathbb{E}[ \| (X+Z) - (Y+Z) \|^p ] = \mathbb{E}[ \| X - Y \|^p ].$$
Taking the infimum over $(X,Y)$ then gives the inequality $d_W(\mu * \nu, \mu' * \nu) \le d_W(\mu, \mu')$.

Note that we used very little--no restrictions on $\nu$, $p$, or even the metric $\|\cdot - \cdot \|$ on $\mathbb{R}^d$ we used except that it was translation invariant. The same argument in fact shows the analoguous statement for any group endowed with an invariant metric (where convolution is defined as the law of a product of independent random variables).

Title: Convolution is a Lipschitz Map


问:
Need help on setting up a double integral over a given region, in polar coordinates
The task is as follows:

Find double integral (in polar coordinates) over region $D$ of $f(x,y)
 = (x^2 + y)$ by $dxdy$.

The region $D$ is bounded by a circle of radius $3$ and a circle of
radius $2$, with upper angle $a1$ being $\frac{\pi}{4}$ and lower angle
$a2$ being $\frac{\pi}{6}$

The region is as the sketch below:

I need help on setting up the double integral.

My work so far:

(1) Let $x = r \cos \theta$ and $y = r \sin \theta$

(2) The double integral in polar coordinates should be something in nature of
$$ \int_{\theta_1} ^ {\theta_2} \int_{r_1}^{r_2} f(r\cos\theta,r\sin\theta) r dr d{\theta}$$

(3) Base on the sketch of the region, $r$ should go from $2$ to $3$,

But I don't see the lower and upper bounds for $\theta$.  Should it be from $0$ to $a1 + a2$?

Or should I consider two integrals instead?  One with $\theta$ from $0$ to $a1$ and another integral with $\theta$ from $0$ to $a2$?

I just want to make sure that my original thoughts are right (or not).

Would someone please help me on this?

Thank you.

答: the angular range will be as given from $-\frac{\pi}{6}$ to $\frac{\pi}{4}$ assuming $a1$ is the angle above the $x-$axis not including the angle $a2$, otherwise the upper value is $\frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}$</p>


答:
Letting $f\colon (0,\infty)\times(0,\infty)$ be defined as
$$
f(x,y) = \sqrt{\log \frac{x^2+y^2}{2xy}}
$$
you question is equivalent to showing that $f$ defines a metric. (This is because we do have, for all $x,y>0$, that $f(x,y)\geq 0$, $f(x,y)=f(y,x)$, and $f(x,y)=0$ iff $x=y$, already).

It remains to prove the triangle inequality, which is the focus of this question. To do so, let us rewrite, for $x,y>0$.
$$
f(x,y) = \sqrt{\log\left(\frac{\frac{x}{y} + \frac{y}{x}}{2}\right)} \tag{1}
$$
and, going further, let us define $h\colon\mathbb{R}\to\mathbb{R}$ by
$$
h(t) = \sqrt{\log\left(\frac{e^t + e^{-t}}{2}\right)} = \sqrt{\log\cosh t}, \qquad t\in\mathbb{R} \tag{2}
$$
so that
$
f(x,y) = h\!\left(\log\frac{x}{y}\right)
$ for all $x,y>0$. Now, the inequality we want to show is
$$
f(x,y) \leq f(x,z) + f(y,z), \qquad x,y,z>0\tag{3}
$$
for which it is sufficient to prove, if I am not mistaken, that
$$
h(s+t) \leq h(s)+h(t), \qquad s,t\in\mathbb{R}\tag{4}
$$
by setting $s=\log\frac{x}{z}$, $t=\log\frac{z}{y}$, so that $s+t=\log\frac{x}{y}$ (which was the whole point of introducing $h$: making things nice and additive).
This last inequality follows from the fact that $h$ is subadditive, because concave (one can differentiate twice, I assume, to see it, and deal with $0$ separately).

Title: Interesting logarithmic inequality

Body:
Recently the following question was published on the site and soon after this deleted:

Prove the inequality
$$
\left(\log\frac{a^2+b^2}{2ab}\right)^n\le \left(\log\frac{a^2+c^2}{2ac}\right)^n+\left(\log\frac{b^2+c^2}{2bc}\right)^n\tag1
$$
for all positive $a,b,c$ and $n=\frac12$.

The inequality has many nice features. Particularly $\frac12$ seems to be the largest possible value of $n$ such that (1) generally holds.

I have succeeded to prove the inequality in a rather awkward way but I believe there should be a simple and nice way to prove it. Possibly it is just a particular case of a more general inequality.

Any hint is appreciated.


问:
Product of two separable spaces separable
Prove that the product of two nonempty separable spaces is separable.

Attempt:Let $X$ and $Y$ be two nonempty separable spaces. Let $A$ be a countable and dense subset of $X$. Let $B$ be a countable and dense subset of $Y$. Since the cartesian product of two countable sets is countable, $A \times B$ is a countable subset of $X \times Y$. Also $\overline {A \times B}=\bar A \times \bar B=X \times Y$ so $A \times B$ is dense.

Comment: I have looked at a proof of this result using projection functions so I am unsure if the strategy I used is correct. Is it really this simple, or is there an issue or flaw with this approach?

答:
There is no problem whatsoever with this approach.You  do have to prove (or know) that $$\overline{A \times B} = \overline{A} \times \overline{B}$$

for all $A \subseteq X, B \subseteq Y$, but after that it's really that simple (for finite products).

There are many ways to approach this problem: another is to use the fact that $D$ is dense iff every basic non-empty open subset of the whole space intersects $D$, applied to basic sets of the form $U \times V$, e.g.


答:
Yes, $(a,x,1)$ and $(b,x,1)$ will be adjacent, assuming that $ab \in E(P_3)$.

It might help to understand what part of this definition makes it lexicographic. As an analogy, consider alphabetizing words in a dictionary:

Which of the following words comes first in the dictionary: "$\text{snowman}$" or "$\text{snowball}$"?

To figure this out, we scan the letters of each word from left to right until we find two letters that don't match. The first four letters are the same, so we look at each word's fifth letter. Since '$\text{b}$' comes before '$\text{m}$' in the alphabet, "$\text{snowball}$" must come before "$\text{snowman}$" in the dictionary. Notice that the remaining letters to the right of these unmatched letters are ignored. Likewise, we can alphabetize the words "$\text{stack}$" and "$\text{exchange}$" by only looking at their first letters (regardless of the remaining letters).

Similarly, to figure out if two vectors of vertices are adjacent, scan along from left to right until you find two vertices that don't match. If these two unmatched vertices are connected by an edge, then the original two vectors of vertices are considered adjacent (regardless of the remaining vertices to the right).

Title: doubt about basic definition

Body:
I have a very basic doubt regarding my last post understanding the basic definiton

Can anybody clear the point what would be the result if I take  $j=1$.

For ex. if I am taking $(a,x,1)$ and $(b,x,1)$ as the vertices, Will they be adjacent or not?
Here the value of $j$ is 1. I made them adjacent as acc to definition...

vertices $(x_1, x_2, . . . , x_k)$, $(y_1, y_2, . . . , y_k)$ are adjacent if for some index j $\in$  {1, 2, . . . , k} we
have $x_jy_j$ $\in$$E(G_j )$ and $x_i = y_i$ for each 1$\leq$ i < j.

In example I took $V(P_3)$ = {a,b,c,d}, $V(K_2)$= {x,y} and $V(K_2)$={1,2}.
And the product is $P_3$$\circ$$K_2$$\circ$$K_2$

Thanks a lot.