问:
Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$
I have tried to wrap my head around this for some time now, and quite frankly I am stuck.

Given is that :
$$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right)$$

Then what is the value of $a/b$?

I tried evaluating the quantities on DESMOS to realise that: a=136 and b=17 which gives 8 as the answer. Any helpful insight about how to reach at these values !?

Body:
$$=∑_{=1}^8 tan^2(\frac{}{17})=136$$

$$b=\prod_{=1}^8 tan^2(\frac{}{17})= 17$$

$$\frac{a}{b}=\frac{136}{17}=8$$

This results can be calculated by hand.

$${Tan[\frac{}{17}]^2, Tan[(2\frac{}{17}]^2, Tan[(3 \frac{}{17}]^2,
 Tan[(4 \frac{}{17}]^2, Cot[(7 \frac{}{34}]^2, Cot[(5 \frac{}{34}]^2,
 Cot[(3 \frac{}{34}]^2, Cot[\frac{}{34}]^2}$$

$${0.0349437, 0.15008, 0.383376, 0.831052, 1.75354, 4.03315, 12.3527, \
116.461}$$

This confirms the formulas by @sangchul-lee: Cauchys proof of the Basel lemma but with the figures calculated.


问:
About possible pathological solutions to the DE $y=y'+y''+y'''+\ldots$
Is it possible to construct (or indirectly show the existence of) a function $y(x)\in\mathcal{C}^{\infty}(\mathbb{R})$ such that series
$$ S(x) = \sum_{n=1}^{+\infty}\frac{d^n y}{dx^n} $$
is pointwise convergent for any $x\in\mathbb{R}$, but $S(x)$ is not a differentiable function?

A more-or-less equivalent and strictly related question is the following one: is it possible for the operator

$$ T:\varphi(x) \mapsto e^x \int_{0}^{x} e^{-t}\varphi(t)\,dt $$
to produce $T(\varphi)\in \mathcal{C}^\infty$ without $\varphi$ being differentiable? Probably no, because...

Assuming that the integral appearing above is the Riemann integral, $T$ acts on the space of almost-everywhere continuous functions, so we are free to assume that $\varphi$ is a.e. $\mathcal{C}^0$, as well as
$$\Phi(x)=\frac{d}{dx}\int_{0}^{x}e^{-t}\varphi(t)\,dt. $$
By the remark in the comments, $\int_{0}^{x}\left(\Phi(t)-e^{-t}\varphi(t)\right)\,dt$ equals zero almost everywhere, so the fundamental theorem of Calculus "almost applies". $T(\varphi)$ is smooth by assumption, so it is $T(\varphi)e^{-x}=\int_{0}^{x}e^{-t}\varphi(t)\,dt$ and its derivative $\Phi(x)$. On the other hand $\Phi(x)=e^{-x}\varphi(x)$ almost everywhere, so $\varphi(x)$ is $\mathcal{C}^\infty$ almost everywhere.

答: If you know that $S(x)$ is continuous, the answer is no. Assume that there is a point $x_{0}\in\mathbb{R}$ such that $S(x_{0})$ is not differentiable. Let $a,b\in\mathbb{R}$ such that $x_{0}\in\left[a,b\right]$. Since $$S(x)=\sum_{k\geq0}y^{(k)}\left(x\right)$$ for every $x\in\left[a,b\right]$ then $y^{(k)}\left(x\right)\rightarrow0$ as $k\rightarrow+\infty$. This means that the formal Taylor series $$\sum_{k\geq0}\frac{y^{(k)}\left(x\right)}{k!}(t-x)^{k}$$ has radius of convergence $\rho(x)=+\infty$, for all $x\in\left[a,b\right]$. In particular there exists some $\delta>0$ such that $\rho(x)\geq\delta$ for all $x\in\left[a,b\right]$ and this is enough to conclude that $y(x)$ is analytic for $x\in\left[a,b\right]$ (see, for example, here). Now, since$$\sup_{x\in\left[a,b\right]}\left|S(x)-\sum_{k=0}^{n}y^{(k)}\left(x\right)\right|=\left|S\left(x^{*}\right)-\sum_{k=0}^{n}y^{(k)}\left(x^{*}\right)\right|\rightarrow0$$ as $k\rightarrow+\infty$ then we can conclude that the convergence is uniform in $\left[a,b\right]$ and so $S(x)$ is analytic for $x\in\left[a,b\right]$; this is a contradiction. Furthermore, we observe that $$\int_{a}^{x}\sum_{k\geq0}y^{(k)}\left(t\right)dt=\int_{a}^{x}y\left(t\right)dt+\int_{a}^{x}\sum_{k\geq1}y^{(k)}\left(t\right)dt=Y(x)+\sum_{k\geq1}y^{(k-1)}\left(x\right)-\sum_{k\geq1}y^{(k-1)}\left(a\right)$$ $$=Y(x)+S(x)-S(a)$$ where $Y(x)=\int_{a}^{x}y\left(t\right)dt$, since the convergence is uniform and so we can exchange the integral with the series. But we also have $$\int_{a}^{x}\sum_{k\geq0}y^{(k)}\left(t\right)dt=\int_{a}^{x}S\left(t\right)dt$$ hence we have the ordinary differential equation of the first order $$S^{\prime}(x)-S(x)=-y(x)$$ and so we have the the family of solutions $$S(x)=e^{-x}\left(c-\int_{a}^{x}y(t)e^{-t}dt\right)$$ for $a\leq x\leq b$.

ContentLicense: CC BY-SA 4.0


问:
In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

What I Tried: Here is the figure :-

I could conclude that $ABCD$ is cyclic, but I really could not use that property in a useful way till now. I thought about Ptolemy's Theorem but I am not sure since I don't know enough lengths.

Instead, Pythagoras Theorem gives the required values, assuming $AD = AB = x$ . But I am still stuck as that is not enough for finding the side-lengths of the quadrilateral, only then I can find it's area. The red angles came to be $45^\circ$, and I am stuck right here. I can't seem to use something with the cyclic quadrilaterals.

Can anyone help me? Thank You.

答:
Some brute force here and not at all elegant w.r.t Calvin Lin's answer..

Join $AC$ and hence see  by cyclic quad.$$\angle ACE=\angle ACB=45^{o}$$ hence $$EC=5,AC=5\sqrt{2}$$ thus $$2x^2={(\sqrt{x^2-25}+5)}^2+BC^2$$ Now apply cosine rule in $\Delta ABC$ to get $$\cos 45=\frac{x^2-50-BC^2}{10\sqrt{2}BC}$$

Solving you can do the rest easily...


问:
What happens to the product before the integral when substitution method?
$$
\int_{r=0}^{r=3/\sqrt2} \int_{\theta=0}^{\theta=2\pi} \left(\sqrt{9-r^2}-r\right) r dr
$$

Then compute them to

$I_1 = 2\pi\int_{r=0}^{3/\sqrt2}\left(\sqrt{9-r^2}\right)rdr$ and $I_2$ which isn't relevant for this question.

When I then use the substitution method I would usually think I would use $u=9-r^2$ and get:

$$2\pi\int_{u=9}^{9/2}\sqrt u.du$$

But when I look In the solution it shows that it becomes

$$-\pi\int_{u=9}^{9/2} \sqrt u.du$$

Could someone explain why the $2\pi$ changes to $-\pi$?
Would be really appreciated.

Body:
$u = 9-r^2, du = -2r \,dr, dr = -\frac{1}{2r}du$

$\displaystyle \int_{r=0}^{r=3/\sqrt2} \int_{\theta=0}^{\theta=2\pi} \sqrt{9-r^2} \, r \, dr \, d\theta = -\pi \int_{9}^{9/2} \sqrt{u} \, du = \pi \int_{9/2}^{9} \sqrt{u} \, du$


问:
Find vectors vertical to given vectors with certain length

Given the vectors $\mathbf{u,v}$ in R³, determine all vectors that are
  vertical to $\mathbf{u}$ and $\mathbf{v}$ with length = 1

Every vector $\mathbf{x'}$ that is to be found must meet these two conditions:

$\mathbf{x'} \cdot \mathbf{u} = \mathbf{x'} \cdot \mathbf{v}=0$

$\mathbf{x'} \times \mathbf{u}= \mathbf{x'} \times \mathbf{v}=1$

My approach is to insert the given vector components into the first condition and get something like this

$
ax_1' + bx_2' + cx_3' =0
$

since the length of the vectors should be 1, there's

$||x'||=\sqrt{( x' \cdot x')}=\sqrt{x_1'^2+x_2'^2+x_3^3}$

is that the right way to go? I don't really know how to tackle this

答:
If you want it to be perpendicular to both $\mathbf{u}$ and $\mathbf{v}$ then take $ \mathbf{u} \times \mathbf{v}$. And then to normalize it (so that it has length 1) just divide this vector by its norm to get

$$\mathbf{x'}=\frac{\mathbf{u} \times \mathbf{v}}{|| \mathbf{u} \times \mathbf{v}||}$$

ContentLicense: CC BY-SA 3.0


问:
Doubt in Problem 6 in Ex 8.2 in Linear Algebra by Hoffman and Kunze
Here is the problem from Ex 8.2 in Hoffman and Kunze

Part $a$, $b$ and $c$ were trivial to solve but I am not able to solve $d$ part. The formula for part $a$ came out to be $$E(x_1,x_2) = \frac{(3x_1 + 4x_2)}{25} (3,4)$$ Now, when I try to solve $d$ part, I thought of simply assuming one basis vector to be $(x_1,x_2)$ and equating to $(1,0)$ but it turns out to be unsolvable. What am I do wrong? Can someone please help? I am reading the book on my own.

答:
For part d let's think about what exactly that means. We want an orthonormal basis $\{v,w\}$ such that $E$ is $\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}$. This is the same thing as saying that $E(v)=v$ and $E(w) = 0$. Well by definition of a projection, anything in the image of $E$ satisfies $E(v)=v$. We want an orthonormal basis, so let's take this to be unit length. For instance, we can take $v=(3/5,4/5)$.

Now we need some nonzero $w$ with $E(w)=0$ and $v\perp w$. That will automatically imply that $v$ and $w$ are linearly independent, giving us our desired basis. Now, $E$ is an orthogonal projection, so its kernel is the orthogonal complement of its image. Thus, we need only find some unit length vector perpindicular to $(3/5,4/5)$. We can take $w=(-4/5,3/5)$.


问:
Making a matrix have a certain rank

Find the values of $a$ and $b$ such that
$$\mbox{rank} \begin{pmatrix}
               3 & 2 & 5 \\
               1 & a & -1\\
               1 & 3 & b\\
              \end{pmatrix} = 2$$

I can only seem to get single values for $a$ and $b$. To get a matrix of rank 2 we need a matrix in row echelon form with one row of $0$'s at the bottom of the matrix. I.e. two linearly independent rows.

I am able to reduce the above matrix to:

$$\begin{pmatrix}
1 & 3 & b\\
0 & -7 & 5-b\\
0 & a-3 & -1-b\\
\end{pmatrix}$$

We get a row of $0$'s at the bottom when $b=-1$ and $a = 3$. Since the questions asks for an equation for a and b I feel like my solution is incomplete. Any advice?

答:
Hint: The determinant of the matrix is
$$
\det(A)=3ab - 5a - 2b + 22.
$$
So one necessary condition for rank two is that $\det(A)=0$. This is an equation in $a$ and $b$ as well.
Indeed, for $(a,b)=(3,-1)$ the determinant is zero.
Can it happen that the matrix has rank zero or one for any $(a,b)$?

Answer: $A$ has rank two if and only if $\det(A)=0$, i.e., $b= \frac{5a-22}{3a-2}$. For $3a-2=0$ the rank is three.

Title: Making a matrix have a certain rank


问:
Can logical equivalence be simplified in terms of simpler logical operators
I have been researching the net for an answer, but sadly to no avail. I know implication can be simplified into simpler operators using de Morgan's and associative laws (such as this post here demonstrates), but I have found no similar solution for equivalence.

The logical compound expression that I am trying to simplify is the following using only conjunction, disjunction and negation:

$$\neg(((A\rightarrow \neg B) \,\&\, C) \vee (\neg A\leftrightarrow\neg C)$$

Thank you very much taken for the explanation!

答:
I suspect you're thinking too hard about this. Look back at how bi-implication is defined in the first place: by definition, $A\leftrightarrow B$ is equivalent to

 $(A\rightarrow B) \,\&\, (B\rightarrow A)$.

So if we know how to "unpack" $\rightarrow$, we can similarly "unpack" $\leftrightarrow$.

Of course this gets lengthy, but that's why we use abbreviations like $\leftrightarrow$ in the first place.


问:
$\int \frac{(e^z)}{z-\pi i} dz$, if C is the ellipse |z - 2| + |z+2| = 6
$$\oint_{C} \frac{e^z}{z-\pi i}\ \mathrm dz$$
If $C$ is the ellipse $|z - 2| + |z+2| = 6$, how is it $0$? I can find the singular point in $2$ and $-2$.

答:
Graph of contour and pole at $z=i\pi$

Perhaps this visual representation can help you intuitively understand why

$$\oint_{C} \frac{e^z}{z-\pi i}\ \mathrm dz=0$$
Where $C$ is the ellipse $|z - 2| + |z+2| = 6$


问:
I can't find the exact limits to find the marginal probability density function (PDF) from a uniformly joint PDF.
For a uniform probability density function (PDF) of two random variables $X$ and $Y$, the PDF exists only in the square of length $\sqrt{2}$, whose vertices are $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$ (please see attached image below, where I also defined the equations of the lines).

Now, we know that the area of this square is $2$, and its PDF is $$f_{X,Y}(x,y)=\frac{1}{2}$$

Now, if I want to find the marginal PDF of $X$, for example, I want to sum the joint PDF over all values $Y$ can take. But I don't know what the limits of integration will be. Do I just sum from -1 to 1 for $y$?

答:
In cases like this it is often handsome to go for finding the CDF of $X$.

Observe that for $x\in[-1,0]$ we have: $$F_X(x)=P(X\leq x)=\frac12(1+x)^2$$

This expression corresponds with area of triangle with vertices $(-1,0)$,$(x,x)$ and $(x,-x)$ divided by $2$.

For $x\in[0,1]$ we find on a similar way that: $$F_X(x)=P(X\leq x)=1-\frac12(1-x)^2$$

To be complete we note that (of course) $F_X(x)=0$ for $x<-1$ and that $F_X(x)=1$ for $x>1$.

Now PDF $f_X(x)$ can be found by differentiating $F_X(x)$ leading to:$$f_X(x)=1-|x|\text{ if }|x|\leq1\text{ and }f_X(x)=0\text{ otherwise}$$


问:
some problems about pigeonhole principle
The Problems

In question A, I know I can set the hole as the picture below. Then I can prove it by pigeonhole principle.

question A

In question B, is it true that I just say that I can place the 16 pieces of kings as the picture below and it is proved?

question B

As for questions C and D, I have no idea how to do next. How should I set the holes? Any suggestion or hint is appreciated. Thanks a lot!

答:
(Fill in the gaps. If you're stuck, show your work and explain where you're stuck.)

D - Do this first (yourself).

It provides a bit of a hint for how to proceed with C, in terms of determining the structure we want to find/avoid. (This is quite common for such setups, where studying the counter example provides some hints.)

C - We want the kings "far apart".

We're told to apply PP. Let's deconstruct the solution (backward thinking from the problem statement).

In particular, $ 5 = \lceil \frac{17}{4} \rceil, 4 = \lceil \frac{16}{4} \rceil$, which suggests that there are 4 holes.

Look at your 16 counterexample. We have 4 holes, each having 4 pigeons.

What could these holes be, to contain kings that definitely do not attack each other?

Hint: 2 kings cannot attack each other if ...

 the squares that they are on have the same parity.

(Note: It's possible that a king cannot attack another king that's in a different hole. We don't care about that.)


问:
Power comparison when alternate hypothesis is true
From what I know, the power of the test is the probability of rejecting $H_0$ when $H_a$ is true. I think I'm getting mixed up with power and $\alpha$, but am I correct in thinking that when $H_a$ is true, we actually want a higher power?

答:
Yes correct.

The power is the probability to Reject $H_0$ when it is false (when is true $H_1$), no error

Type I  error, $\alpha$ is the probabilty to reject $H_0$ when it is true (error)

The probability to Accept $H_0$ when the alternative is true (complement of the power) is "Type II Error"


问:
How do I find the lengths of the two legs of a right triangle if one leg is half the size of the other and I'm given the hypotenuse?
I've tried squaring the hypotenuse and then dividing up that number into 3 parts. I then give 2 of those parts to the bigger leg and one to the smaller leg. Afterward, I find the square root of each part of the hypotenuse squared. Whenever I do this, I end up with two numbers, but neither of them are half of the other.

(P.S. I'm sorry if this question is confusing to read.) (I'm not sure if I put down the right tag either. This is my first time on this website.)

答:
Let the hypotenuse be given as $h$.  Let the short leg be $x$ and the long leg be $2x$.  By the Pythagorean theorem,
\begin{align}x^2+(2x)^2&=h^2\\x^2+4x^2&=h^2\\5x^2&=h^2\\x^2&=\frac{h^2}{5}\\x&=\frac h{\sqrt 5}\end{align}


问:
Algebraic topology book after John Lee "Introduction to topological manifolds"
What algebraic topology book do you suggest to study after one has studied John Lee's "Introduction to topological manifolds"?</p>

答:
I also recently finished Lee's book (although I don't believe one can ever "finish" a math textbook) and I'm finding  J.P May's A Concise Course in Algebraic Topology very illuminating

(it's also available for free on his website which is nice).

When I have trouble understanding some passages or I find the motivation for some things lacking (since May's book is indeed quite concise) I usually consult Hatcher who Is significantly more wordy (in my opinion) and is, as I understand it, the standard reference for algebraic topology

(+ his book can also be found for free on his website). \

I've also found Pierre Schapira's Lecture notes helpful. He is, like May, a bit heavier on the use of categorical terminology (although I haven't read much thus far, but the little I did get around to read seemed great).

Hope this helps :)


问:
Prove the determinant is $0$
This is Problem 16.17 from the book Exercises in Algebra by A. I. Kostrikin.

Prove that
$$
\left|\begin{array}{ccccc}
\dfrac{1}{2 !} & \dfrac{1}{3 !} & \dfrac{1}{4 !} & \cdots & \dfrac{1}{(2 k+2) !} \\
1 & \dfrac{1}{2 !} & \dfrac{1}{3 !} & \cdots & \dfrac{1}{(2 k+1) !} \\
0 & 1 & \dfrac{1}{2 !} & \cdots & \dfrac{1}{(2 k) !} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & \dfrac{1}{2 !}
\end{array}\right|=0, \quad k \in \mathbb{N}
$$

My Attempt: I tried to expand it by the first column, but it seemed to be more complicated when I did that. I also tried to add edges to the determinant(in the hope that it will be easier to calculate), but I still failed to work it out.

So, My Question is, how to calculate this determinant?

答:
Finally I found a direct way to work this out, that is, to use Generating Functions.

From this article we get the following proposition:

Proposition. Consider the following infinite matrix with $1$s in the super diagonal.
$$
D=\left[\begin{array}{ccccc}
b_{0} & 1 & 0 & 0 & \cdot \\
b_{1} & c_{1} & 1 & 0 & . \\
b_{2} & c_{2} & c_{1} & 1 & . \\
b_{3} & c_{3} & c_{2} & c_{1} & . \\
b_{4} & c_{4} & c_{3} & c_{2} & . \\
\cdot & . & . & . & .
\end{array}\right]
$$
Let $B(x)=\sum_{n=0}^{\infty} b_{n} x^{n},$ and $C(x)=\sum_{n=1}^{\infty} c_{n} x^{n}$ be the generating functions for the sequences $b_{0}, b_{1}, b_{2}, \ldots$ and $c_{1}, c_{2}, c_{3}, \ldots,$ respectively. If
$$
A(x)=\frac{B(x)}{1+C(x)}=\sum_{n=0}^{\infty} a_{n+1} x^{n}
$$
then $a_{n}=(-1)^{n-1} D_{n}$ and $1+x A(-x)$ is the generating function of $D_{n}$.

For the original problem, it is equivalent to calculate the determinant:
$$
D=\left| \begin{matrix}
 \dfrac{1}{2!}&  1&  0&  \cdots&  0\\
 \dfrac{1}{3!}&  \dfrac{1}{2!}&  1&  \cdots&  0\\
 \dfrac{1}{4!}&  \dfrac{1}{3!}&  \dfrac{1}{2!}&  \cdots&  0\\
 \vdots&  \vdots&  \vdots&  \ddots&  \vdots\\
 \dfrac{1}{\left( 2k+2 \right) !}&  \dfrac{1}{\left( 2k+1 \right) !}&  \dfrac{1}{\left( 2k \right) !}&  \cdots&  \dfrac{1}{2!}\\
\end{matrix} \right|
$$

We can see that $\displaystyle B(x)=\sum _{i=0}^{\infty } \frac{x^{i}}{(i+2)!}= \frac{-x+e^x-1}{x^2}$, $\displaystyle C(x)=\sum _{i=1}^{\infty } \frac{x^i}{(i+1)!} = \frac{-x+e^x-1}{x}$, so $A(x)=\dfrac{1}{x}+\dfrac{1}{1-e^x}$, and finally the generating function for $D_n$ is $D(x)=\dfrac{x}{e^x-1}+x$.

Notice the order $N$ of original matrix $D$ is always odd (since $N=(2k+2)-2+1=2k+3$), so the original claim is equivalent to when $n>4$ and $n$ is odd, $[x^n]D(x)=0$. It's easy to see that $D(x) -1 - x/2$ is an even function, which implies its series only contains terms of the form $x^{2k}$, thus we finished our proof.

I think this method really requires a good understanding of linear equations and Cramer's Law, along with the knowledge of generating functions.

Title: Prove the determinant is $0$


问:
Show that the only solution to $\phi(n) =n-2$ is $n=4$
Came across this question in Number Theory.

Let $\phi$ denote Euler's totient function;  Show that the only
  solution to $\phi(n) =n-2$ is $n=4$

My workings so far have included, firstly proving that $4$ is indeed a solution, I have also noted that the solution cannot be prime because $\phi(p)=p-1$.

From here, could I note that my solution must be composite, and the only way in which it could have only 2 numbers which are not coprime would be if n is a square number?

Not sure how to prove this completely though.

Also, could I incorporate the fact that  $\phi(p^k)=p^{k-1}(p-1)$ in any way?

All help greatly appreciated. Thanks.

答: $\varphi(n)$ is even for every $n \ge 3$, so you would need $n$ to be even as well. However, if you write $n = 2^k m$, $m$ odd, then $$ n - 2 = \varphi(n) = \varphi(2^k) \varphi(m) = 2^{k-1} \varphi(m) \le 2^{k-1} m = \frac{n}{2}$$ which severely limits the possibilities for $n$.</p>


问:
Characterizing graphs for which the subdivision graph S(G) is Eulerian or Hamiltonian
I know that $G$ is Eulerian iff all of the vertex degrees are even. So my thinking is that for any cycle graph $C_n$, its subdivision graph is Eulerian because each vertex has degree 2 and adding a new vertex at each edge doesn't change that. But I think my professor wants a more comprehensive answer. For the bipartite graph $K_{n,m}$ the subdivision graph will be Eulerian if $n$ and $m$ are both even so that the degree of all vertices is even.

From this, is it safe to say that the requirement for $S(G)$ being Eulerian is that $G$ is Eulerian? Am I missing something?

Would it also be correct to say that the requirement for $S(G)$ being Hamiltonian is that $G$ is Hamiltonian?

答:
I'm going to assume $S(G)$ is $G$ with every edge subdivided. Let $|G| = |V(G)|$ and $||G|| = |E(G)|$.

$G$ is Eulerian if and only if $S(G)$ is Eulerian.

Subdividing an edge adds a vertex of degree $2$ and preserves all other degrees, and thus if $G$ is Eulerian $S(G)$ is Eulerian. If $S(G)$ is Eulerian, suppose $G$ is not Eulerian. Find a contradiction.

If $G$ is Hamiltonian (has a Hamiltonian cycle), then it is not necessarily the case that
$S(G)$ is Hamiltonian.

Can you find infinitely many examples? Here's one: $K_4$.

In order to characterize you need to make a few observations.

Suppose $S(G)$ is Hamiltonian and let $C$ be a Hamiltonian cycle in $S(G)$. Note, $S(G)$ replaces each edge $uv$ of $G$ with a vertex $a$, and adds edges $ua,av$. Let $A\subseteq S(G)$ be the set of all such $a$. For every $a \in A$, we have $\deg_{S(G)}(a) = 2$. We know $C$ must use $|S(G)|$ edges. Simultaneously, $C$ must use every edge with an end point in $A$ (why?). Conclude from this that $$|S(G)| = |C| = ||S(G)||$$ and thus $S(G)$ is a cycle, and thus $G$ is a cycle.

For your last question, if I understand it correctly.

If $S(G)$ is Hamiltonian, then $G$ is Hamiltonian.

Hint: Find a Hamiltonian cycle in $S(G)$, how can you extend this to a Hamiltonian cycle in $G$.


问:
Proving inner product conditions.
I'm new to the concept of inner product space and was wondering if anyone cold help me here.

I believe that the inner product space is the multiplication of each corresponding entry of the two vectors and then the addition of all of these values.

So I believe to show part (a) ii., I have to say that for any length of $\vec v$ you will always have $0* \vec v_{n} =0$ and that $0+0+0+0...$ n times will always result in 0.

However for part (a) i., I don't know how to approach this at all.

Am I headed on the right track or is my thinking incorrect?

Any help would be appreciated.

答:
for part (a) i., I don't know how to approach this at all

For (a) i., we have $\langle \vec u,\vec v\rangle=\vec u^T\vec v$, so $\langle M\vec u,\vec v\rangle=(M\vec u)^T\vec v.$

By a property of transpose, $(M\vec u)^T=\vec u^TM^T$.

Therefore $\langle M\vec u,\vec v\rangle= \vec u^TM^T\vec v=\langle \vec u,M^T\vec v\rangle.$

Title: Proving inner product conditions.


问:
Is forcing needed to prove existence of a model of ZF that is a countable stage of L?
[Edit]: This is an edit of question [the original question ends before the context is added], it only added the general context of that question, the question itself remains the same.

Is forcing needed to prove that there is a model of ZF that is a countable stage of Gödel's constructible universe? Formally, does the following requires forcing to prove it?

$$\exists \alpha : |\alpha|=\aleph_0 \land L_\alpha \models ZF$$

The context for that question is the following:

First I'll quote some of Noah's answer to a prior posting:

Quoting from this answer:

"we can build the $L$-hierarchy on any "starting set" of Quine atoms $A$ - let $L^A_0=A$, $L^A_\lambda=\bigcup_{\alpha<\lambda} L^A_\alpha$ for $\lambda$ limit, and $L^A_{\alpha+1}=L^A_\alpha\cup\mathcal{P}_{def}(L^A_\alpha)$. Letting $W_A$ be the $L$-hierarchy on $A$ continued until we get a model of ZF-Reg, we can show that the pure part of $W_A$ is countable regardless of whether $A$ is; now simply take an uncountable set of Quine atoms".

This was used to show that there can exist a set in ZF-Reg. that is not subnumerous to any well founded set; thereby proving that both choice and CH can fail in some models of ZF-Reg. [since $A$ won't be comparable to its Hartog number]. The same argument can be adopted to prove that failure in some models of ZFA.

Now the line written in bold is exactly what I'm asking about. My understanding of that is that the pure part of $W_A$ cannot be countable unless its height is countable, i.e. the $\alpha$ such that $W_A = L^A_\alpha$, must be countable. Now according to the answer suppose that $A$ is empty, then the "pure part" of $W_A$ would be $W_A$ itself, which mean that we can have a model of ZF at some countable stage of L.

The broader context of this question is the following:

Suppose that the proof of this countability doesn't require forcing. Now instead of letting $A$ be a set of quine atoms, let $A= L_{\omega+1} \cup P(\omega) $ [or one can let $A=V_\alpha$] of some transitive model $M$ of ZF, then build up a construction over $A$ along the above lines, i.e. iterate $A$ in a constructible manner up to some countable height so that $W_A \models ZF$. Now by the same argument we'd reach to the conclusion that $|P(\omega)| \neq \omega_1$, since $P(\omega)$ would be seen by $W_A$ to be not comparable to its Hartog number! The reason is because $P(\omega)$ is seen by $M$ to be uncountable, while each ordinal in $W_A$ would be seen by $M$ to be countable, so there is no injection in $W_A$ from $P(\omega)$ to its Hartog, while $\omega_1$ equals its Hartog is something provable in $W_A$, so $W_A$ doesn't satisfy the continuum hypothesis.

I realize that there must be something wrong in that line of thinking, otherwise this would mean that forcing is not needed to prove independence of CH and choice from ZF.

答:
Forcing is totally irrelevant to the question of whether there is a countable level of $L$ satisfying $\mathsf{ZFC}$.

First, note that since the $L$-hierarchy is absolute and forcing adds no new ordinals, we have that forcing cannot change whether a model thinks there is a level of $L$ satisfying $\mathsf{ZFC}$: if $M[G]$ is a generic extension of $M$, then $$M\models\exists\alpha(L_\alpha\models\mathsf{ZFC})\quad\iff\quad M[G]\models\exists \alpha(L_\alpha\models\mathsf{ZFC}).$$

Next, we bring this down to the countable. Suppose $L_\alpha\models\mathsf{ZFC}$; applying downward Lowenheim-Skolem and then condensation, we get that there is some countable $\beta$ such that $L_\beta\models \mathsf{ZFC}$. So even though forcing can make uncountable things countable, it cannot affect the fact that the first level of $L$ satisfying $\mathsf{ZFC}$ - if such exists at all - is countable. Note moreover that this entire argument goes through in $\mathsf{ZFC}$ (or indeed vastly less).

(Relatedly, my quoted claim was a bit imprecise, albeit in a standard way: what we can show is that if $W_A$ exists at all then it has countable height and countable pure part. This is a genuine $\mathsf{ZFC}$ theorem, whereas what I wrote is an extension of $\mathsf{ZFC}$ - $\mathsf{ZFC}$ + "$\mathsf{ZFC}$ has transitive models of arbitrarily large cardinality" is enough.)

What about the putative $\mathsf{CH}$-related argument?

Well, the issue is that the result above simply doesn't apply to the situation you're considering. It's extremely limited to the specific "$L$-on-$A$" hierarchy defined above.

When we build the $L$-hierarchy on $A$ - in which context I'll write "$L_{\alpha,A}$" as opposed to "$L_\alpha^A$," since the latter clearly clashes with relativization notation (no idea what I was thinking originally) - it's crucial that there be no "structure" in the set of atoms $A$. From the perspective of each $L_{\alpha,A}$ (including $W_A$) the atoms are totally indistinguishable from each other: the subsets of $A$ in $L_{\alpha,A}$ are exactly the finite or cofinite ones, and (externally) every permutation of $A$ extends to a (unique) automorphism of $L_{\alpha,A}$. This means that we can't in any meaningful sense take $A$ to be $L_{\omega+1}\cup\mathcal{P}(\omega)$.

Of course this motivates an interesting question:

Suppose we have a structure $\mathfrak{X}=(X; R_1,...,R_n)$ (for simplicity I'm assuming the language is finite - if we want to use infinite languages, some care must be taken). What is the smallest height $h_\mathfrak{X}$ of a well-founded model $M$ of $\mathsf{ZF-Reg}$ with set of atoms $A$ such that there are relations $S_1,...,S_n\in M$ on $A$ with $$(A; S_1,...,S_n)\cong (X; R_1,...,R_n)$$ (externally of course)?

That is, how tall must a model of $\mathsf{ZF-Reg}$ "built on a copy of $\mathfrak{X}$" be? I don't know much about this specific question, but if we replace $\mathsf{ZF-Reg}$ with $\mathsf{KPU}$ we get a very well-trod topic in higher recursion theory - the canonical starting point for this is Barwise's wonderful (and freely-available!) book Admissible sets and structures.

Finally, given this question I should probably sketch the proof of my claim (granting the above-mentioned hypothesis) in the first place.

There are a couple ways to prove it, but in my opinion the "right" proof is via the following:

$(*)\quad$ Suppose $A,B$ are infinite sets of Quine atoms. Then for each ordinal $\alpha$ we have $$L_{\alpha,A}\equiv L_{\alpha,B}.$$

$(*)$ can be proved via Ehrenfeucht-Fraisse games. It implies my original claim as follows. Suppose $A$ is uncountable. Let $B$ be a countably infinite subset of $A$. By a downward Lowenheim-Skolem + collapsing argument, we have that $height(W_B)$ is countable. But by $(*)$ we have that $$L_{height(W_B), A}\models\mathsf{ZFC}.$$ So we have $height(W_A)\le height(W_B)<\omega_1$.

EDIT: if you prefer forcing to Ehrenfeucht-Fraisse games, here's another proof of $(*)$:

Consider a generic extension $V[G]$ of the universe in which both $A$ and $B$ are countable. In $V[G]$ we have  $L_{\alpha,A}\cong L_{\alpha,B}$ (build an isomorphism between the two recursively from a bijection $A\rightarrow B$). A fortiori we have in $V[G]$ that $L_{\alpha,A}\equiv L_{\alpha,B}$. But first-order theories don't change under forcing, and so we get $L_{\alpha,A}\equiv L_{\alpha,B}$ in reality.

(This sort of trick can be used for all sorts of shenanigans. For example, one way to prove that every satisfiable $\mathcal{L}_{\omega_1,\omega}$-sentence $\varphi$ has a countable model is to collapse the cardinality of a given $\mathcal{M}\models\varphi$ to $\aleph_0$. We wind up with a generic extension satisfying the $\Sigma^1_1$ sentence "$\varphi$ has a countable model," so by Mostowski absoluteness $\varphi$ had a countable model in the ground universe already.)


问:
$x^3 + y^3 = p^2$ over the integers
$x^3 + y^3 = p^2$ has a solution over the integers for some three digit prime p. Find all p that satisfy.

The first thing I did was factorize the left hand side, getting $(x+y)(x^2 - xy + y^2) = p^2$

I then considered the case $x^2 - xy + y^2 = p^2$, which gave me $p^2 = 3y^2 -3y +1$, which can be naturally factorized into $(p+1)(p-1) = 3y(y-1)$, but how do I do anything with this? I don't see a continuation from this point onwards.

Any hints or solutions would be greatly appreciated. Thanks!

答:
Consider the linear factor first instead; this allows you to directly express $y$ in terms of $x$. From
$$p^2=x^3+y^3=(x+y)(x^2-xy+y^2),$$
it follows that $x+y=p^k$ for some $k\in\{0,1,2\}$, and $x^2-xy+y^2=p^{2-k}$. Then $y=p^k-x$, so we can eliminate $y$ to get the quadratic equation
$$3x^2-3p^kx+p^{2k}-p^{2-k}=0.$$
A quadratic has an integral root if and only if its discriminant is a perfect square. Can you continue from here?

 The discriminant equals $$\Delta=(-3p^k)^2-4\cdot3\cdot(p^{2k}-p^{2-k})=-3p^{2k}+12p^{2-k}.$$ For $k=1$ and $k=2$ this becomes $$\Delta=-3p^2+12p=-3p(p-4)\qquad\text{ and }\qquad\Delta=-3p^4+12,$$ which are both negative because $p>4$. Hence $k=0$ and $\Delta=12p^2-3$. So if $p$ is such a prime, then $\Delta$ is a perfect square and a multiple of $3$, say $\Delta=(3e)^2$. Then a bit of algebra shows that $(X,Y)=(2p,e)$ is an integral solution to the Pell equation $$X^2-3Y^2=1.$$

Title: $x^3 + y^3 = p^2$ over the integers


问:
Upper bound of an operation
Let $L$ be the following operation
$$L: = I - \Delta, $$
from $C(R^2)$ to $C(R)$, where $I$ is the identity operation and $\Delta$ is the Laplacian operation. From the equation $Lu = f$, where $f$ is a known function, we have
$$u = {L^{ - 1}}f.$$
By taking the norm, from both sides of the last equation, we have
$${\left\| u \right\|_p} = {\left\| {{L^{ - 1}}f} \right\|_p} \leqslant {\left\| {{L^{ - 1}}} \right\|_p}{\left\| f \right\|_p}$$
Now, I need an upper bound for ${\left\| {{L^{ - 1}}} \right\|_p}$.
I will appreciate it if someone helps me to find that upper bound.

答:
Here is a heat kernel approach: Consider the heat semigroup $(e^{t\Delta})$. It is well known (and can be seen quite easily using the Fourier transform) that
$$
e^{t\Delta}f(x)=(4\pi t)^{-n/2}\int_{\mathbb{R}^n}e^{-\frac{|x-y|^2}{4t}}f(y)\,dy.
$$
Direct computation shows that $\|e^{t\Delta}f\|_p\leq \|f\|_p$. In fact, it suffices to show this for $p=2$ and $p=\infty$; the other cases follow by interpolation and duality.

The resolvent is the Laplace transform of the heat semigroup, that is,
$$
(-\Delta+\lambda)^{-1}f=\int_0^\infty e^{-t\lambda}e^{t\Delta}f\,dt.
$$
Thus
$$
\|(-\Delta+\lambda)^{-1}f\|_p\leq \int_0^\infty e^{-t\lambda}\|e^{t\Delta}f\|_p\,dt\leq \|f\|_p\int_0^\infty e^{-t\lambda}\,dt=\frac 1{\lambda}\|f\|_p.
$$
In your particular case you get $\|L^{-1}\|_{p\to p}\leq 1$.

Title: Upper bound of an operation


答:
Here is why your argument is incomplete. One way for your reduced matrix to have rank 2 is to have row three all zeros, as you said. HOWEVER, there is another possibility: row 2 and 3 being multiple of each other. (With row 1 this is not a question b/c row 1 cannot be multiple of either row 2 or row 3 since we have 1 versus 0s on first column.

So let us look at this second possibility: we will have rank 2 if
$$
\frac{-7}{a-3} = \frac{5-b}{-1-b} \, .
$$
Which hopefully is the same equation as the one in the answer below reached by computing determinant.

Title: Making a matrix have a certain rank

Body:
Find the values of $a$ and $b$ such that
$$\mbox{rank} \begin{pmatrix}
               3 & 2 & 5 \\
               1 & a & -1\\
               1 & 3 & b\\
              \end{pmatrix} = 2$$

I can only seem to get single values for $a$ and $b$. To get a matrix of rank 2 we need a matrix in row echelon form with one row of $0$'s at the bottom of the matrix. I.e. two linearly independent rows.

I am able to reduce the above matrix to:

$$\begin{pmatrix}
1 & 3 & b\\
0 & -7 & 5-b\\
0 & a-3 & -1-b\\
\end{pmatrix}$$

We get a row of $0$'s at the bottom when $b=-1$ and $a = 3$. Since the questions asks for an equation for a and b I feel like my solution is incomplete. Any advice?


问:
Convergence of improper integral $\int_0^1 t^p \sin t dt$
Need help with this problem. For $p\in \mathbb{R}$ consider the improper integral
$$I_p=\int_0^1 t^p \sin t dt$$
Which of the following is statements are true regarding the convergence of $I_p$?

A. $I_p$ is convergent for $p=-1/2$

B. $I_p$ is divergent for $p=-3/2$

C. $I_p$ is convergent for $p=-4/3$

D. $I_p$ is divergent for $p=-1/2$

First of all, I could not understand why is it improper. Secondly the answer is provided as all correct. But could not get explanation for that either. Can you guide me please ? Tried to search in MSE to get any answer for that, but in vain. If anywhere it is showed how to solve, please help me to get the link. I will study my self and try to understand.

答:
For $-2 < p <0$, using $\sin t \leq t$, we have
\begin{align*}
I_p & = \int_0^1 t^p \sin t \ dt \\
& \leq \int_0^1 t^{1+ p}  \ dt \\
& \leq \frac{1}{p + 2} < \infty.
\end{align*}
Hence, the integral converges for $-2 < p < 0$. For $p \leq -2$, first note that $\sin t \geq t/2$ for $t \in [0,1]$. Hence,
\begin{align*}
I_p & = \int_0^1 t^p \sin t \ dt \\
& \geq \frac{1}{2}\int_0^1 t^{(1+ p)}  \ dt.
\end{align*}
Since the integral on RHS diverges for $p \leq -2$, $I_p$ also diverges.


问:
If $\angle p + \angle q + \angle r + \angle s + \angle t = 500^\circ$, find $\angle A + \angle B + \angle C + \angle D + \angle E$.

Consider the following star figure. If $\angle p + \angle q + \angle r + \angle s + \angle t = 500^\circ$, find $\angle A + \angle B + \angle C + \angle D + \angle E$.

What I Tried: Here is the figure :-

Some might already know the sum of the interior angles of an $n$-angled star, is $(n - 4)180^\circ$. Using that one might not deduce that fast that the answer is not $180^\circ$ in this case. In fact, the star is defined such that we draw it from a specific point, and continue making it without lifting the pencil.

For this, this is not the case. In fact it is more difficult to answer this question, from my point of view. The answer was given as $140^\circ$ . Joining lines and taking  variables will take a long time, and in some cases the extended lines will not meet at good places, making it more hard.

Can anyone help me how is that the answer?

答:
Considering the angle sum of the star figure, which is actually a 10-gon:

$$(360^\circ - p)+\dots + (360^\circ -t)+\angle A + \dots + \angle E = (10-2)\times 180^\circ $$

$$360^\circ \times 5 - (p+q+r+s+t)+\angle A + \dots + \angle E = 8\times 180^\circ $$

Now solve for $\angle A + \dots + \angle E$.


问:
Embedding and existentially quantified formula
I wanted to prove that given an existentially quantified formula $\varphi(\bar{y})=\exists\bar{x}\psi(\bar{x}, \bar{y})$ and an embedding $f : \mathfrak{M} \hookrightarrow \mathfrak{N}$ between two $\mathscr{L}$-structures $\mathfrak{M}$ and $\mathfrak{N}$, there is for any $\bar{a} \in M^k$,
$$\mathfrak{M} \models \varphi(\bar{a}) \Rightarrow \mathfrak{N} \models \varphi(f(\bar{a})). \tag{*}$$

($f(\bar{a})$ denotes $f(a_1), \ldots, f(a_n)$).

I already proven that given a quantifier-free formula $\psi$,  $$\mathfrak{M} \models \psi(\bar{a}) \iff \mathfrak{N} \models \varphi(f(\bar{a})) \tag{**}.$$ I have a wrong proof for (*) but wasn't able to figure out wherever was erroneous:

$$
\begin{align*}
\mathfrak{M} \models \exists \bar{x} \psi(\bar{x}, \bar{a}) &\iff \textrm{exists $\bar{b} \in M^k$, s.t. $\mathfrak{M} \models \psi(\bar{b}, \bar{a})$} \tag{Definition}\\
&\iff \textrm{exists $f(\bar{b}) \in N^k$, s.t. }\mathfrak{N} \models \psi(f(\bar{b}), f(\bar{a})) \tag{**}\\
& \iff \mathfrak{N} \models \exists\bar{x} \psi(\bar{x}, f(\bar{a})) \tag{Definition}
\end{align*}
$$

I'm not sure which logical equivalence is wrong.

答:
First let's show that the converse $$\mathfrak{N}\models\varphi(f(\overline{a}))\quad\implies\quad\mathfrak{M}\models\varphi(\overline{a})$$ is false in general; this will help uncover the issue with your proof.

Working in the empty language for simplicity, let $\mathfrak{M}$ be a one-element structure and let $\mathfrak{N}$ be a two-element structure. Let $a$ be the unique element of $\mathfrak{M}$ and fix some $f:\mathfrak{M}\rightarrow\mathfrak{N}$; note that any such $f$ is an embedding. Then we have $$\mathfrak{N}\models\exists x(x\not=f(a))\quad\mbox{but}\quad \mathfrak{M}\models\neg\exists x(x\not=a).$$ So $\exists$-sentences are only "upwards absolute" in general.

OK, so based on this where is the issue in your argument?

Well, the problem above was that the witness to the existential statement holding in $\mathfrak{N}$ was not itself in the image of $f$, so it couldn't be "pulled down" to a witness in $\mathfrak{M}$. This shows us the problem in reversing the final implication:

 We do have $$\mathfrak{N}\models\varphi(f(\overline{a}))\quad\implies\quad \exists \overline{c}\in\mathfrak{N}\mbox{ such that }\mathfrak{N}\models\psi(\overline{c},f(\overline{a})),$$ but we cannot assume that there is $\overline{b}\in\mathfrak{M}$ such that $f(\overline{b})=\overline{c}$. So we do not have $$\mathfrak{N}\models\varphi(f(\overline{a}))\quad\implies\quad \exists \overline{b}\in\mathfrak{M}\mbox{ such that }\mathfrak{N}\models\psi(f(\overline{b}),f(\overline{a}))$$ in general.


问:
Asymptotic bound for $\int_0^\infty \int_0^\infty (x+y)^m e^{-\frac{x^2}{2i} - \frac{y^2}{2j}} dx\, dy\;$ for $i$ and $j$ large
Whilst trying to count certain types of bipartite graphs, I'm lead to try to bound the following quantity
$$
I:=\int_0^\infty \int_0^\infty (x+y)^m e^{-\frac{x^2}{2i} - \frac{y^2}{2j}} dx\,dy
$$
where $i,j$ and $m$ are integers, and I'm interested in the asymptotics for large $i$ and $j$ and potentially $m$ (although it would suffice to have a good upper bound when $i \approx j$ and $m=o(i)$).

One can derive an exact expression for the integral by multiplying out the terms and using known identities for the quantities $\int_0^\infty x^k e^{-\frac{x^2}{2i}} dx$, however the asymptotics of this sum is unclear to me.

It would seem more natural to use a type of `saddle-point' method here, approximate the logarithm of the function around its maximum at $(x_0,y_0) = \left(i \sqrt{\frac{m}{i+j}},j \sqrt{\frac{m}{i+j}} \right)$ using the first two terms of the Taylor series, and so evaluate the integral in this region as a standard Gaussian, and then show that the contribution from outside this region is negligible.

This would lead to the following bound, which I would guess is in fact the correct asymptotic order
$$
I \approx \exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)\pi\sqrt{2ij}.
$$
However, I can't get the regions in which the approximation is correct and the region in which the integral is negligible to overlap.

I suspect that this integral will have been considered somewhere in the literature, or at the very least will be susceptible to standard techniques in a field I'm not familiar with.

答:
$$I=$$

For $i\inℕ$ and $j\inℕ$ always true. But $y\inℝ$ and $x\inℝ$ without restrictions.

If the restriction are presented first to the integral then the solutions looks this way:
$$I(x/;ℝ,y/;ℝ,i/;ℕ,j/;ℕ,m/;ℕ)=$$

This is done with Mathematica and Wolfram Language. I used the built-ins $Assuming$ and $Integrate$. This should be working with CAS that have not the high-level mathmatical knwoledge implemented.


问:
Is every positive definite matrix also positive semidefinite?
I am in trouble with the definitions of positive definite and positive  semidefinite matrices. By definition, does the following implication hold?

$$\mbox{positive definite} \implies \mbox{positive semidefinite}$$

I guess yes. Look at the following example. Let

$$A = \begin{vmatrix}
4 &-2& 0\\
-2& 4& -2\\
0&-2&4
\end{vmatrix}$$

It is a diagonally dominant matrix $(|a_{ii}|\geq \sum_{j\neq i}|a_{ij}|$ for all $i$), so it is semidefinite positive. But, if I compute the eigenvalues, they are all strictly positive which implies actually that it is definite positive.

My question is: everytime I find that a matrix is semidefinite positive, thus I have to use another criterion in order to try to understand if it is "less" that semidefinite positive (i.e. definite positive)? Or am I wrong with something?

Thank you in advance!

答:
Positive definite means "All eigenvalues are greater than zero."

Positive semi-definite means "All eigenvalues are greater than or equal to zero."

The first of these implies the second.

The second does not imply the first, as the all-zero matrix shows.


问:
Consequences of negating Russell's infinity axiom in FOL
Apologies in advance for another curious question.

From my research, it is clear that for any natural number we can construct a model that would entail a sentence such that there is the same number of elements in the domain of that model.
So, why not generalize this by using the negation of Russell's infinity axiom as a sufficient means to achieve the same ends (i.e. the entailment of any sentence in a finite domain)?

答:
It sounds like you're looking for a sentence which is satisfied in exactly the finite structures. By the compactness theorem$^1$, no such sentence exists in first-order logic. The crucial point is that a structure may be "wrong about itself." The classic example of this is the existence of nonstandard models of first-order Peano arithmetic: these are discrete ordered semirings which satisfy the first-order induction scheme, so "think" that all their initial segments are finite, yet "from the outside" have infinite elements. A first-order "axiom of finity" may rule out obvious infiniteness-es, but it won't be able to genuinely ensure that the structure in question is finite.

Of course this is a limitation of first-order logic specifically. In second-order logic for example we can indeed write down a sentence true in exactly the finite structures. But there are good technical reasons for wanting to stick to first-order logic; ultimately compactness (and its "cousin in limitation," the Lowenheim-Skolem theorem) is a good thing to have.

$^1$Specifically, suppose $\varphi$ is a first-order sentence true in structures of arbitrarily large cardinality. Then the theory $$\{\varphi\}\cup\{\exists x_1,...,x_n(\bigwedge_{1\le i<j\le n}x_i\not=x_j): n\in\mathbb{N}\}$$ is finitely satisfiable, hence by compactness has a model. But such a model must be infinite.


问:
Check whether the following series is uniformly convergent OR not?
Check whether the following series is uniformly convergent  OR not?

$\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}, x\in [0,\infty).$

My attempt

$\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}=x^4\sum_{n=1}^\infty \frac{1}{(1+x^2+x^4)^n}=( \frac{1}{(1+x^2+x^4)}).(\frac{x^4}{1-(\frac{1}{1+x^2+x^4})})=\frac{x^4}{(x^2+x^4)}$

$|\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}-\frac{x^4}{(x^2+x^4)}.1|=|\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}-\frac{x^4}{(x^2+x^4)}.\sum_{n=1}^\infty \frac{1}{2^n}|=|\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}-\sum_{n=1}^\infty \frac{x^4}{(x^2+x^4)}. \frac{1}{2^n}|=|\sum_{n=1}^\infty (\frac{x^4}{(1+x^2+x^4)^n}-\frac{x^4}{(x^2+x^4)}. \frac{1}{2^n})|$
How do I proceed further?
Is there any theorem helps to solve it faster?
I couldn't apply weirstrauss M-Test.

答:
By generalizing your first line, one can show the tail sum equals
$$\sum_{n=N+1}^\infty \frac{x^4}{(1+x^2+x^4)^n} = \frac{x^4}{(1+x^2+x^4)^{N+1}}\frac{1}{1-\frac{1}{1+x^2+x^4}}
= \frac{x^2}{1+x^2} \cdot \frac{1}{(1+x^2+x^4)^N}.$$

To show uniform convergence, it suffices to check that the following holds.
$$\sup_{x \ge 0} \left(\frac{x^2}{1+x^2} \cdot \frac{1}{(1+x^2+x^4)^N}\right) \overset{N \to \infty}{\longrightarrow} 0.$$


问:
Find the flux of the vector field $F(x,y,z)=(x+y^2, -x, z-xy)$

Let $S$ be a surface given by $x^2+3y^2+z^2=1$ and the vector field
$F(x,y,z)=(x+y^2, -x, z-xy)$.

Find $\iint _{S} F \cdot \vec{n}$
$dS$ with $\vec{n}$ pointing outside of the surface.

I've tried finding this using Gauss' theorem as the divergence of the vector field is an scalar ($div F = 2$).

So: $\iint_S F dS = \iiint_V 2 dV$

Now I made a change of variables with the surface, with $\begin{cases} x=u \\ y=\frac{v}{\sqrt3} \\ z=w \end{cases}$ then I have $u^2+v^2+w^2=1$, with the Jacobian $J=\frac{1}{\sqrt 3}$.

And then I made another change of variables but with spherical coordinates:

$\begin{cases} u=\rho \cos \theta \sin \phi \\ v = \rho \sin \theta \sin \phi \\ w = \rho \cos \phi \end{cases}$

with $\rho \in [0,1]$, $\theta \in [0, \pi]$, $\phi \in [0, 2\pi]$, and the Jacobian $J=\rho^2\sin \theta$.

The integral I got was $\int _{0} ^{1} \int _{0} ^{2\pi} \int _{0} ^{\pi} \frac{2}{\sqrt3} \rho^2 \sin \theta d \phi d\theta d\rho = 0$, and that's not the correct answer...

 The correct answer is $\iint _{S} F \cdot \vec{n} = \frac{8\pi}{3 \sqrt 3}$

答:
Your integral order is wrong.

$\displaystyle \int _{0} ^{1} \int _{0} ^{2\pi} \int _{0} ^{\pi} \frac{2}{\sqrt3} \rho^2 \sin \theta \, d \phi \, d\theta \, d\rho = 0$

Based on notation you have used, $\theta$ is the polar angle and $\phi$ is the azimuthal angle.

So, $0 \leq \theta \leq \pi \,$ and $0 \leq \phi \leq 2\pi \,$

Your integral should be

$\displaystyle \int _{0} ^{1} \frac{2}{\sqrt3} \rho^2 \bigg[ \int _{0} ^{2\pi} \bigg [\int _{0} ^{\pi} \sin \theta \, d \theta\bigg] \, d\phi \bigg] \, d\rho$

$\,$

You get $ \,\displaystyle \int _{0} ^{1} \int _{0} ^{2\pi} \int _{0} ^{\pi} \frac{2}{\sqrt3} \rho^2 \sin \theta \, \color{blue} {d \theta \, d\phi} \, d\rho = \frac{8 \pi}{3\sqrt3}$


问:
Is $\sum_{p=0}^{k}(-1)^p\binom{k}{p}p^k=\sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+a)^k, a \in \mathbb{R}$ true? If so, how?
I come here from this answer actually: https://math.stackexchange.com/a/3743805/765852

Due to re-indexing,
$$\sum_{p=1}^{k+1}(-1)^p\binom{k}{p-1}p^{k+1}$$
should equal
$$-\sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+1)^k,$$
yet it's written as
$$-\sum_{p=0}^{k}(-1)^p\binom{k}{p}p^k.$$ I tested some results with my calculator and it seems that this is not only true for $(p+1)^k$, but also for $(p+a)^k$, where $a \in \mathbb{R}$. Is this the case? If so, how would one go about proving this?

I actually tried my hand at proving this (it's just using simple algebra). However it requires a property which I also tried proving but I am not sure about its proof. I'm going to shamelessly plug that in here (it's open): How would I prove $\sum_{p=1}^{n}\binom{n}{p}p^m(-1)^p=0, m \in \mathbb{N}, 1\leq m < n $ via induction?

Ok here's the proof:

Statement:
$$\sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+a)^k=\sum_{p=0}^{k}(-1)^p\binom{k}{ p}p^k, a \in \mathbb{R}$$

Proof:
\begin{align*}
 \sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+a)^k &= \sum_{p=0}^{k}(-1)^p\binom{k}{p}(p^k+kp^{k-1}a+\binom{k}{2}p^{k-2}a^2+...kpa^{k-1}+a^k) \\
      &= \sum_{p=0}^{k}(-1)^p \binom{k}{p}p^k+ (-1)^p\binom{k}{p}kp^{k-1}a + (-1)^p\binom{k}{p} \binom{k}{2}p^{k-2}a^2... \\
      &= \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^k+ \sum_{p=0}^{k} (-1)^p\binom{k}{p}kp^{k-1}a+ \sum_{p=0}^{k} (-1)^p\binom{k}{p} \binom{k}{2}p^{k-2}a^2... \\
      &= \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^k+ka \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^{k-1}+a^2 \binom{k}{2} \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^{k-2}...
        \end{align*}

Then, by the property linked above, it should be
$$\sum_{p=0}^{k} (-1)^p\binom{k}{p}p^k+0+0+....+0+ \sum_{p=0}^{k} (-1)^p\binom{k}{p}$$

Simply by the binomial theorem, the last term should also be zero. This completes the proof.

答:
For any function $f$ defined on the real numbers, let $\def\D{\Delta}\D f$ denote the function defined by
$$
(\D f)(x)=f(x+1)-f(x)
$$
You can then iterate this construction. Let $\D^n f$ denote $\D(\D(\dots(\D f)\dots))$, with $n$ applications of $\D$. For example,
\begin{align}
(\D^2f)(x)
&=(\D f)(x+1)-(\D f)(x)
\\&=f(x+2)-2f(x+1)+f(x)\\
(\D^3 f)(x)&=f(x+3)-3f(x+2)+3f(x+1)-f(x)\\
(\D^4 f)(x)&=f(x+4)-4f(x+3)+6f(x+2)-4f(x+1)+f(x)
\end{align}
and so on. You might see Pascal's triangle emerging, and indeed, it is easy to prove by induction that
$$
(\D^k f)(x)=\sum_{p=0}^n (-1)^{k-p}\binom{k}pf(x+p)
$$
Now, letting $f(x)=x^k$, and factoring out $(-1)^k$, we get that
$$
(\D^k f)(x)=(-1)^p\sum_{p=0}^k (-1)^p\binom{k}p(p+x)^k\tag1
$$
You want to prove that the value of the last equation is independent of $x$. This is equivalent to saying that $\D^k f$ is a constant function. This is proven using the following observation:

For any polynomial $f$ of degree $n$, where $n>0$, $\D f$ is a polynomial of degree $n-1$.

Therefore, since $f(x)=x^k$ has degree $k$, it follows $\D^k f$ has degree zero, so is constant, so $(1)$ is independent of $x$.


问:
Series convergence proof and determining limit value
I want to show that the following series is convergent and determine the corresponding limit value.

$$ a(n):=\sum_{k=1}^n \frac{k}{n^2}$$

To show the series converges I used the ratio test:

$$\left|\frac{k}{(n+1)^2}\right|\cdot\left|\frac{n}{k}\right|=\frac{n}{(n+1)^2}<1$$

So the series converges. But how can I go on now and calculate the limit value?

Body: $$\sum _{k=1}^n \frac{k}{n^2}=\frac{1}{n^2}\sum _{k=1}^nk=\frac{1}{n^2}\frac{n(n+1)}{2}=\frac{n+1}{2n}\to\frac12$$


问:
Volume of solid of revolution around a line
What is the volume of the solid that results when the region bounded by the graphs
of $y=\sqrt{x}$, y = 0, and x=4 is revolved about the line x = 4.

My solution is
$$ \pi \int_0^2 (4-y^2)^2 \, dy = \frac{256\pi}{15} $$

I don't know if this correct. How can I integrate this in terms of x? In terms of x, is the result the same with my integration above?

答:
Integrate over x with the variable change $y=\sqrt{x}$

$$ \pi \int_0^2 (4-y^2)^2 \, dy =
\pi \int_0^4 (4-x)^2 \frac{dx}{2\sqrt{x}}\\ =
\pi \int_0^4 \left(\frac{8}{\sqrt{x}}- 4{\sqrt{x}}
+ \frac12x\sqrt{x}\right)dx=
 \frac{256\pi}{15} $$


问:
Tensor product of maps between Hilbert spaces
Let $H_1,H_2, K_1,K_2$ be Hilbert spaces. Let $S_1: H_1 \to K_1$ and $S_2: H_2 \to K_2$ be bounded linear maps. Is it true that there exists a bounded linear map
$$S_1 \otimes S_2: H_1 \otimes H_2 \to K_1 \otimes K_2$$ such that $(S_1 \otimes S_2) (\xi_1 \otimes \xi_2) = (S_1 \xi_1) \otimes (S_2 \xi_2)$? Here the tensor products are not algebraic tensor products! Rather, they are completions of the algebraic tensor product.

I know this result is true when $H_1 = K_1$ and $H_2 = K_2$, but is this true as well?

答:
Embed each of your spaces as a closed subspace of a bigger space, say $H_1\subset\tilde H_1$, etc., such that $$\dim\tilde H_1=\dim\tilde K_1,\qquad \dim\tilde H_2=\dim\tilde K_2.$$ You can trivially extend $S_1$ and $S_2$ to operators $\tilde S_1,\tilde S_2$ by making them $0$ on $H_1^\perp$ and $H_2^\perp$ respectively.  Let $U:\tilde H_1\to\tilde K_1$, $V:\tilde H_2\to\tilde K_2$ be unitaries. Now you can use the existence of a bounded linear map
$$
U^*\tilde S_1\otimes V^*\tilde S_2:\tilde H_1\otimes \tilde H_2\to \tilde H_1\otimes \tilde H_2
$$
that acts on elementary tensors as you want. Now note that for unitaries, it is trivial to check that $U\otimes V$ is isometric on sums of elementary tensors, which means that $U\otimes V$ is a well-defined unitary on the tensor product. So we can define
$$
S_1\otimes S_2=(U\otimes V)(U^*\tilde S_1\otimes V^*\tilde S_2)|_{H_1\otimes H_2}.
$$
This map is bounded, being a composition of bounded operators, and
$$
(S_1\otimes S_2)(\xi_1\otimes \xi_2)=(U\otimes V)(U^*\tilde S_1\xi_1\otimes V^*\tilde S_2\xi_2)
=S_1\xi_1\otimes S_2\xi_2.
$$


问:
Why does $a^{\log_a(x)}=x$?
I'm currently learning about logs and yesterday made a post on here.

In that post I was told that $a^{\log_a(x)}=x$

From a commenter:

"It should be clear that..."

Nope, not for me. I've tried coming back to this since posting yesterday but cannot 'see it' or make it click. I'm seeking hand holding and a low level response. Why does $a^{log_a(x)}=x$?

答:
It is true that by definition $a^{\log_a(x)}=x$. However, perhaps it would be worthwhile to show you the motivation behind why we define logarithms in this way. As @Stinking Bishop has already pointed out, logarithms help us solve for $x$ when dealing with an equation of the form
$$
a^x=b \, .
$$
For example, if
$$
10^x=100
$$
then $x=2$ is the solution. But suppose we were trying to solve
$$
10^x=101 \, .
$$
Now, it is not immediately clear what $x$ is. We can approximate its decimal value using trial and error, but this is only an approximation. In the same way that $\sqrt{2}$ represents the exact positive solution of $x^2=2$,
$$
\log_{10}(101)
$$
represents the exact solution to the equation
$$
10^x=101 \, .
$$
This means that
$$
10^{\log_{10}(101)}=101
$$
by definition. $\log_{10}(101)$ is the 'label' we give to the solution of
$$
10^x=101 \, .
$$
More generally, if $a^x=b$, then $\log_a{b}$, by virtue of how logarithms are defined, represents the solution of the equation. Thus,
$$
a^{\log_a{b}}=b
$$
by definition. Now, a more sophisticated way of thinking about logarithms is that they inverse exponentiation. If $f(x)=a^x$, then $f^{-1}(x)=\log_a(x)$. The defining feature of inverse functions is that $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$ for all $x$. Hence,
$$
\log_a(a^x) = x \text{ and } a^{\log_a{x}}=x
$$
are both true by definition. The hardest part is trying to explain why these two conceptions of logarithms, while superficially different, are actually the same. Let's return to our equation
$$
a^x=b \, .
$$
If we 'take logs of both sides', we get
$$
\log_a(a^x) = \log_a(b) \, ,
$$
which simplifies to
$$
x= \log_a(b).
$$
Thus, by defining the logarithm as the inverse of the exponential, $x$ represents the power that we must raise $a$ to in order to get $b$. And this should make intuitive sense. In general, the inverse function should tell us 'how to go backwards'. If we start with the number $x$ and perform a function on it so that we get $f(x)$, then $f^{-1}$ is like an instruction manual for how to get back to $x$. Thus,
$$
f^{-1}(f(x))=x \, .
$$
In the case of exponentiation, we start with a number with a number $x$, and raise $a$ to the power of $x$ to get $a^x$. The logarithm, being the inverse function, should let us return to $x$:
$$
\log_a(a^x)=x \, .
$$
Also, if we first perform $f^{-1}$ on $x$, then again, $f$ is the inverse function. Try working out what this means in the context of exponentials and logarithms.

Title: Why does $a^{\log_a(x)}=x$?


问:
What does an increase in the mean, but the median is the same signify?
I have been collecting some data from a project (check if something is affected by the environment it is in) using some sensors. I collected some data before making a change in the environment and then after. My hypothesis is that it should be affected by the environment (the number must decrease).

To test this, I did some calculations (in Excel) and noticed that there is a small increase in the mean, but the median remains the same. I don't know how to interpret this (I only have some basic knowledge about statistics). Can I simply assume that there was indeed an increase (and that my hypothesis is false), or is there something more to it? After searching online, I found something about called "hypothesis testing of means" that seems to be applicable here, but I have to idea how to do this (tried watching videos about it, but didn't understand how to use it in my case). I calculated the variance and the standard deviation if it helps.  Thanks.

Here is the result of my data:

Sample size: 60

BEFORE:

mean: 3.35

median: 3.5

variance: 0.25

standard dev.: 0.50

AFTER:

mean: 3.51

median: 3.5

variance: 0.20

standard dev.: 0.45

答:
A single new large data point will increase the mean but leave the median the same. Think about what happens to the salary distribution in a company if the CEO gets a giant raise.

To know the "meaning" of your observation we need a lot more context. An Excel plot of the data should help you.

That said, this is a better question for stats.stackexchange.com.


问:
Evaluate $\int \frac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$
I`m trying to solve this integral and I did the following steps to solve it but don't know how to continue.
$$\int \dfrac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$$

$$\begin{align}\int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{\sin^2(x)}\,\operatorname d\!x &= \int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{1-\cos^2(x)} \\
&=\int \sin^{-2}(x)\,\operatorname d\!x + \int \cos(x)\,\operatorname d\!x - \int \frac{\operatorname d\!x}{\cos(x)}\end{align}$$
Any suggestions how to continue?

Thanks!

Body:
Hints:

$$(1)\;\;\int\frac{1+\cos x}{\sin^2x}dx=\int\frac1{\sin^2x}dx+\int\frac{\cos x}{\sin^2x}dx$$

$$(2)\;\;\;\;\;\;(\cot x)'=-\frac1{\sin^2x}$$

$$(3)\;\;\;\text{for a derivable function}\;f : \int\frac{f'(x)}{f(x)^m}dx=\frac{f(x)^{1-m}}{1-m}+C\;,\;\;\color{blue}{m\neq -1}$$


问:
Obtaining coefficients of a sum of exponential decays
Let $f:\mathbb R^+\to\mathbb R^+$ be such that
\begin{equation}
f\left(x\right)=\sum_{n=0}^\infty a_ne^{-4\pi nx}
\end{equation}
where $a_n$ are nonnegative integers. Is there some way of recovering $a_n$ from $f$? I have been trying to get them like coefficients of a Fourier series, and have tried taking a Laplace transform. Maybe something else works.

答:
You actually can do the same thing you'd do for a Fourier series. The general theory is that if we have a Hilbert space with the inner product $\langle\cdot\vert\cdot\rangle$ and an orthonormal basis $\{b_i\}_{i\in\mathbb N}$, then for any element
$$v=\sum_{k=0}^\infty a_k b_k$$
of the Hilbert space we can recover the coefficients $a_n$ by calculating the inner product of $v$ and $b_n$, since we have
$$\langle v\vert b_n\rangle=\left\langle \sum_{k=0}^\infty a_kb_k\middle\vert b_n\right\rangle=\sum_{k=0}^\infty a_k\langle b_k\vert b_n\rangle=a_n,$$

where the second equality comes from the continuity and linearity of the inner product, the third equality comes from orthonormality of the base.

This is what we do when calculating Fourier coefficients. The Hilbert space is that of functions with a given period (let's say $1$). The orthonormal basis is the set of functions $\exp\left(2\pi\mathrm inx\right)$, and the inner product wrt which it is orthonormal is
$$\langle f\vert g\rangle:=\int_0^1 f(x)\cdot g^\ast(x)\mathrm dx,$$
where $X$ is the set on which the functions are defined. Calculating the Fourier coefficients is basically calculating $\langle f\vert\exp(2\pi\mathrm inx)\rangle$. We can do the same with functions of the form
$$\sum_{k=0}^\infty a_n\mathrm e^{-4\pi nx}.$$
We have to find an inner product on this space with respect to which the given basis functions $\mathrm e^{-4\pi nx}$ are an orthonormal system. I give you this one:
$$\langle f\vert g\rangle:=\int_0^1 f\left(\frac{\mathrm ix}{2}\right)g^\ast\left(\frac{\mathrm ix}{2}\right)\mathrm dx.$$

You can show that it has all the necessary features of an inner product, and that $\langle\exp(4\pi nx)\vert\exp(4\pi kx)\rangle$ with respect to this inner product is the same as $\langle\exp(2\pi\mathrm i nx)\vert\exp(2\pi\mathrm ikx)\rangle$ with respect to the one used for calculating Fourier coefficients. Calculating these "Laplace coefficients" (I don't think that's an official term) works the same as calculating Fourier coefficients. Just that with the given functions, you only have to index them with non-negative integers.


问:
Why is copula entropy negative?
Most statistical measures are non-negative. Shannon entropy, $H(X)$, a statistical measure of disorder or uncertainty in probability distributions, is also non-negative, despite it having a negative sign at the front due to logarithm rules. (Probability values are non-negative too.)

The entropy of the copula density, $c(u,v)$, however, is negative, despite it having a negative sign at the front like Shannon entropy. The negative coefficient actually makes the positive value in the double integral negative in the end.
$$h(c(u,v)) = -\iint_{[0,1]^2} c(u,v) \log c(u,v) \, \mathrm{d}u \, \mathrm{d}v$$

Why is it negative? and what is the intuition then when we see, for example, a Gaussian copula density with an entropy of -3.5? Is it more uncertain than a copula entropy of -2.1?

From the opposite perspective, why isn't copula entropy positive, or wouldn't it be better if copula entropy were always positive?

Ma, J. and Sun, Z. (2011), 'Mutual information is copula entropy',
Tsinghua Science and Technology. 16(1), 51-54.

答:
Most statistical measures are non-negative.

Not true for differential entropies.

The "entropy of copula density" is negative just like the (differential) entropy of any density restricted to $[0,1]$ is negative.

Title: Why is copula entropy negative?


问:
Let $ABCDEF$ be a hexagon such that the diagonals $AD,BE,CF$ intersect at point $O$.

Let $ABCDEF$ be a hexagon such that the diagonals $AD,BE,CF$ intersect at point $O$, and the area of the triangle formed by any three adjacent points is $2$. Find the area of the hexagon.

What I Tried: Here is a picture :-

I drew some kind of an irregular hexagon for convenience, but I am thinking that is this true only for a certain regular hexagon? In this case the triangles are definitely don't have equal area as $2$, is it possible only for a regular hexagon, along with diagonals $AD,BE,CF$ concurring at $O$? If it is true (which I am not sure), then it is great, but I have found no way to prove it.

As another way, by adding and subtracting areas, I was only able to conclude that $[\Delta BFD] = [\Delta AEC]$, other than that I have no idea.

Can anyone help me here? Thank You.

答:
Given that the area of the triangle formed by any three adjacent points are equal, there are a lot of parallel lines in this hexagon. For example:

$$[\triangle ABC] = [\triangle BCD] \implies BC//AD \implies [\triangle BOC] = [\triangle ABC] = [\triangle BCD]$$

Using this property for each side of the hexagon, the area is $2 \times 6$.

EDIT: Found a squished hexagon with concurrent diagonals which are parallel to the sides:


问:
Proving equality of sup/inf given a function $r:X \times Y\longrightarrow [0,\infty]$
Given a function $r\colon X\times Y \longrightarrow [0,\infty]$, and $\alpha\in[0,\infty]$, we define the following relation: $$ (x,y)\in r_{\alpha}\Longleftrightarrow r(x,y)\leq\alpha.$$

Let $\mathscr{U}$ be an ultrafilter on $X$ and $\mathscr{V}$ an ultrafilter on $Y$. Then I'm trying to prove the following equality: $$\sup_{U\in\mathscr{U},V\in\mathscr{V}}\ \inf_{x\in U, y\in V} r(x,y) = \inf\big\{\alpha\in[0,\infty]\ \big| r_{\alpha}\mathscr{U}\subseteq\mathscr{V}\big\}. $$

Any help will be appreciated!

答:
I think I found a solution to this question.

First, assume $r_{\gamma}\mathscr{U}\subseteq\mathscr{V}$. Take an arbitrary $A\in\mathscr{U}$ and $B\in\mathscr{V}$. Then one has that $r_{\gamma}A\in\mathscr{V}$ and hence there is some $y\in r_{\gamma}A\cap B$. Choose $x\in A$ such that $r(x,y)\leq\gamma$. In particular, one has
$$\inf_{x\in A,y\in B}r(x,y)\leq\gamma,$$
so that $\sup_{A\in\mathscr{U},B\in\mathscr{V}}\inf_{x\in A,y\in B}r(x,y)$ is indeed a lower bound for $\{\gamma\in[0,\infty] | r_{\gamma}\mathscr{U}\subseteq\mathscr{V}\}$.

To prove the other inequality, assume $\gamma<\inf\{\alpha\in[O,\infty] | r_{\gamma}\mathscr{U}\subseteq\mathscr{V}\}$. Hence, $r_{\gamma}\mathscr{U}\not\subseteq\mathscr{V}$ and thus there is an $A\in\mathscr{U}$ so that $r_{\gamma}A\notin\mathscr{V}$. This means there must be a $B\in\mathscr{V}$ for which one has $r_{\gamma}A\cap B=\emptyset$. Take now an arbitrary $x\in A$ and $y\in B$. Since $y\notin r_{\gamma}A$, one has
$$ r(x,y)\geq\gamma.$$
This shows that $\gamma\leq\sup_{A\in\mathscr{U},B\in\mathscr{V}}\inf_{x\in A,y\in B}r(x,y)$, which in turn proves (by the randomness of $\gamma$) the remaining inequality.

Is this a correct proof?


问:
Integrating $\frac{1}{(1-x)^2} $ into two different-looking functions
Background

I want to "trick" some students by showing that $$\int \frac{1}{(1-x)^2} \mathrm dx = \frac1{1-x} + C$$ in one instance and $$\int \frac{1}{(1-x)^2} \mathrm dx = \frac x{1-x} + C$$ in another.

Obviously these look like two different results due to the different numerator, but as the more astute will point out, the difference between these functions is a constant, and they therefore have the same derivative.

Question

When solving the integral, using the substitution $u = 1-x$ we naturally arrive that the first results, but is there a way of solving the integral that "naturally" yields the second result?

答:
Why not get them to differentiate $$\frac{x}{1-x}$$ using the quotient rule:

$$\frac{\mathrm d}{\mathrm dx}\left(\frac{x}{1-x}\right) = \frac{1.(1-x) - (-1).x}{(1-x)^2}$$ which simplifies very naturally to give$$\frac{1}{(1-x)^2}.$$

This would imply $$\int \frac{1}{(1-x)^2} \mathrm dx = \frac x{1-x} + C$$


问:
Show that $\angle PAB = \angle CAQ$
Given a triangle $\triangle ABC$ with a point $P$ in the perpendicular bisector of $BC$. Take point $Q$ also in the bisector of $BC$ s.t. $\angle ABP + \angle ACQ =180^o$ and such that $P$ and $Q$ are in the interior of $\angle BAC$.

Prove that $\angle PAB = \angle CAQ $

I tried to find some circles in this configuration. Tried to find some interesting locus and I failed so I went for trigonometry. This problem is equivalent to show that $\frac{BP}{AP} = \frac{BQ}{AQ}$... wait this looks like an apolonian circle.

Ok it looks like $CQ$ is tangent to the apollonian circle that goes throught $P$ and $Q$ EDIT: it is not tangent.

Can we work it with coaxial systems? Tha apollonian circle through fixed points $AB$ form a coaxial system orthogonal to the system of circles that pass throught $A$ and $B$

答:
We use @Edward Porcella's diagram!

Let $PQ\cap\odot(ABC):=\{G,H\}$ and hence,
$$ \angle ABP+\angle ACQ = 180=\underbrace{\angle ACG}_{\angle ACQ-\angle GCQ} +\underbrace{\angle ABG}_{\angle ABP+\angle PBG}$$$$\implies \angle PBG=\angle GCQ=\angle GBQ $$
Hence, $BG$ and $BH$ are internal and external angle bisectors of $\angle QBP\implies \odot(ABC)$ is $B$-Apollonian Circle of $\triangle PBQ$ and as $A$ lies on this circle, we get $AG$ bisects $\angle QAP$ and we are done!$$\tag*{$\blacksquare$}$$

Title: Show that $\angle PAB = \angle CAQ$


答:
Put $x=\sin^2(t)$; then $dx = 2\sin(t)\cos(t)\,dt$. We have
$$
\int \frac{1}{(1-x)^2}\,dx \longrightarrow \int \frac{2\sin(t)\cos(t)}{(1-\sin^2(t))^2}\,dt
$$
$$
=2\int \frac{\sin(t)}{\cos^3(t)}\,dt = 2\int \tan(t)\sec^2(t)\,dt
$$
$$
\begin{cases}\stackrel{y=\tan(t)}{\longrightarrow}\int 2y\,dy= \tan^2(t)+C = \frac{\sin^2(t)}{1-\sin^2(t)}+C\longrightarrow \frac{x}{1-x}+C\\
\stackrel{y=\sec(t)}{\longrightarrow}\int 2y\,dy=  \sec^2(t)+C = \frac{1}{1-\sin^2(t)}+C\longrightarrow \frac{1}{1-x}+C
\end{cases}
$$This is a general phenomenon: antiderivatives of trig functions involving tangent and secant often look dissimilar but by the FTC they must be identical up to a constant.

Title: Integrating $\frac{1}{(1-x)^2} $ into two different-looking functions

Body:
Background

I want to "trick" some students by showing that $$\int \frac{1}{(1-x)^2} \mathrm dx = \frac1{1-x} + C$$ in one instance and $$\int \frac{1}{(1-x)^2} \mathrm dx = \frac x{1-x} + C$$ in another.

Obviously these look like two different results due to the different numerator, but as the more astute will point out, the difference between these functions is a constant, and they therefore have the same derivative.

Question

When solving the integral, using the substitution $u = 1-x$ we naturally arrive that the first results, but is there a way of solving the integral that "naturally" yields the second result?


问:
Is there an easy way to compare floors and ceilings? Eg, is it true that $\lfloor{\frac{n}{k}}\rfloor\leq\lceil{\frac{n+1}2}\rceil$ for $k\geq 2$?
I want to see if for all $n, k \in \mathbb{Z}, k \geq 2$,
$$\left\lfloor{\frac{n}{k}}\right\rfloor \leq \left\lceil{\frac{n+1}{2}}\right\rceil$$
I've checked a bunch of numbers with some basic python, but how would I show this mathematically?

Edit: I have a lot more inequalities like this to show, so I'd love to also know the method you would use. Thanks!

答:
The statement does not hold... Even if $n,k \in\mathbb{N}$, in the case $k=1$ you would have to show that
$$
n \leq \left\lceil \frac{n+1}{2} \right\rceil
$$

and this is just nor true. If $n,k \ge 2$ the result holds. To check that, just need to show that it holds for $k=2$, because for larger $k$ the LHS is even smaller.

Considering any $k \ge 2$, the result does not hold for in general for negative $n$. For instance, if $k\ge 4$, the result only holds for $n\ge -2$.


答:
Consider the statement $$\color{red}{2}^\color{green}{3}=\color{blue}{8}$$

You can say that:

$\color{green}{3}=\log_\color{red}{2}(\color{blue}{8})$

$\color{red}{2}=\sqrt[\color{green}{3}]{\color{blue}{8}}$

Instead of asking "what number must $\color{red}{2}$ be raised to, to give $\color{blue}{8}$" you may ask "what is $\log_\color{red}{2}(\color{blue}{8})$?". You can then consider $$\color{red}{a}^{\log_\color{red}{a}(\color{blue}{x})}$$

$\log_\color{red}{a}(\color{blue}{x})$ is "the number that gives $\color{blue}{x}$ when $\color{red}{a}$ is raised to it", so Очевидно, что $$\color{red}{a}^{\log_\color{red}{a}(\color{blue}{x})}=\color{blue}{x}$$

In general (given that both $\log_a(c), \sqrt[b]c$ exist) if $$\color{red}{a}^\color{green}{b}=\color{blue}{c}$$ then

$\color{green}{b}=\log_\color{red}{a}(\color{blue}{c})$, which means that $$\color{red}{a}^{\log_\color{red}{a}(\color{blue}{c})}=\color{blue}{c}$$

$\color{red}{a}=\sqrt[\color{green}{b}]{\color{blue}{c}}$, which means that $$(\sqrt[\color{green}{b}]{\color{blue}{c}})^{\color{green}{b}}=\color{blue}{c}$$

I recommend watching this.

Title: Why does $a^{\log_a(x)}=x$?

Body:
I'm currently learning about logs and yesterday made a post on here.

In that post I was told that $a^{\log_a(x)}=x$

From a commenter:

"It should be clear that..."

Nope, not for me. I've tried coming back to this since posting yesterday but cannot 'see it' or make it click. I'm seeking hand holding and a low level response. Why does $a^{log_a(x)}=x$?


问:
Representatives of a collection of sets
Suppose that a collection of sets $A_{1},A_{2},\dots,A_{n}$ satisfies $|\bigcup\limits_{j=1}^{k}A_{{i}_{j}}|\geq k+1$ for every $1\leq k <n$ and $\{i_1,i_2,\dots,i_k\} \subset [n]$, and that $x \in A_i$.

Show that the collection has a system of distinct representatives in which $x$ represents $A_i$.

It was suggested to me that graphs may be useful in solving this but I can't see how.

答:
Consider the bipartite graph $G = (S,T,E)$, where the vertices in $S = \{v_1,\dots,v_n\}$ correspond to the sets $A_1,\dots,A_n$, the vertices in $T = \{u_y : y \in \cup A_j\}$ correspond to the elements of $\cup A_j$, and edges denote set membership, so that $u_yv_j$ is an edge precisely if $y \in A_j$.

In this language, what you would like to show is that there is a matching covering $S$ which contains the edge $u_xv_i$. In other words, there is a matching covering $S-v_i$ in $G - v_i - u_x$. Now you just have to decipher what the condition on the sizes of the $A_i$ say for this graph and use a well-known result about matchings in bipartite graphs (which hopefully you are familiar with).

Title: Representatives of a collection of sets


问:
A smooth family of symplectic forms
Let $A(t)\in\mathbb R^{2n\times 2n}$ be a smooth family of nondegenerate skew-symmetric matrices, $t\in\mathbb R$. Then the family $A(t)$ represents the family of symplectic forms $\omega_t(u,v)=\langle A_tu,v\rangle$. Is it possible to construct a smooth family $S(t)=\{u_1(t),\ldots,u_n(t),v_1(t),\ldots,v_n(t)\}$ of bases in $\mathbb R^{2n}$ such that $$\omega_t(u_i(t),u_j(t))=\omega_t(v_i(t),v_j(t))=0,\ \ \ \ \ \ \ \omega_t(u_i(t),v_j(t))=\delta_{ij}$$ for all $t\in\mathbb R$? It's possible to construct $S(t)$ in a small neighborhood of every $t_0\in\mathbb R$, but I can't do this for all $t$. Thanks for helping.</p>

答:
Yes, we can do that.

Let $S$ be the set consists of all non-degenerated skew symmetric matrices in $Mn(\mathbb{R},2n)$. And define $p:Gl(\mathbb{R},2n)\to S$ by $p(X)=X^{T}JX$, where J is the following matrix:
\begin{pmatrix}
0 & -I_n\\
I_n & 0\\
\end{pmatrix}
then by Proposition 4.3 of Helgason's "Differential Geometry, Lie groups, and Symmetric Spaces", we have a diffeomorphism $\tilde{p}:Sp(\mathbb{R},2n)\backslash Gl(\mathbb{R},2n) \to S$.

Then your question is reformulated as lifting a smooth curve in $S$ to $Gl(\mathbb{R},2n)$. And we find it's trivial to do it locally by the structure of homogeneous space. Then use Lebesgue method on the integral where the curve is defined, we can glue them up.

Title: A smooth family of symplectic forms


问:
Inequality for $p$ vector norm
Is it true for $2 \leq q \leq p$ that
$$
\|x\|_q\leq n^{\frac{p-q}{pq}} \|x\|_p
$$
where $x$ is an $n$-dimensional vector. I only need the inequality for $n=2$, so that would suffice. I'm just curious if it's true for any $n$. Just from plugging it into a graphing calculator I believe it to be true, but how would I prove it?

答:
$$
\lVert x\rVert_q^{p }=\left(\frac 1n\sum_{i=1}^n\lvert x_i\rvert^q\right)^{p/q}n^{p/q}
$$
and using convexity of the map $t\mapsto t^{p/q}$, one gets
$$
\lVert x\rVert_q^{p }\leqslant  \frac 1n\sum_{i=1}^n\lvert x_i\rvert^p  n^{p/q}
$$
or in other words,
$$
\lVert x\rVert_q^{p }\leqslant n^{p/q-1}\lVert x\rVert_p^p.
$$

Title: Inequality for $p$ vector norm


问:
The subspace of eventually zero sequences. Banach algebra.
I'm trying to figure out if the space $c_{00}=\{(x_n)|\exists N\:  \forall n>N \: x_n=0\}$ is a hyperplane and the maximum ideal in the space $c_0=\{(x_n)|\lim x_n=0\}$.

I understand that the closure $c_{00}$ will be $c_0$. But that doesn't help me solve the problem.

I tried to consider the factor $c_0 / c_{00}$ and it is the space of sequences converging to zero with different tails. But $c_0$ itself does not have a unital element in it either. Is this a sufficient reason to say that then $c_0 / c_{00}$ is not a field and $c_{00}$ is not the maximum ideal.

Perhaps I should consider some functionality with the $c_{00}$ kernel?

Help please with the idea of the solution.

答: I guess $c_{0 0}$ is not a hyperplane. If $X = \{\frac{1}{n}\}_{n \geq 1}\,\bigcup\,\{0\}$ is equipped with the normal metric in $\mathbb{R}$ then $c = \{\{x_n\}\,\vert\,\lim_{n \rightarrow \infty} x_n$ exits$\}$ is actually the entire $C(X)$. In this case $c_0$ is a hyperplane in $c$ but meanwhile $c_{0 0}$ is dense (in $\|\cdot\|_{\infty}$) in $c_0$. If there is a bounded linear functional defined on $c_0$ with kernel $c_{0 0}$, then we can extend that functional to $c$ but then that extension will have kernel equal to $c_0$. Alternatively you can use the fact that $c_0^{\ast} = \ell_1$ and hence no bounded linear functional will have its kernel equal to $c_{0 0}$


问:
Proving an equality of functors in algebraic theories
Can somebody explain how are they claiming the following highlighted identity in the book Algebraic Theories by Vitale?

The reason I am confused is because the image of the $F$ functor lands inside $\mathcal{B}$ whereas the image of the $\mathcal{B}(-, FX)$ functor lands inside the category whose objects are morphism classes in $\mathcal{B}$.

答:
It certainly is just a misprint.
After removing the '$={\cal B}(-,FX)$' part one obtains

'[...] because for $A=Y_{{\cal C}^{op}}(X)={\cal C}(-,X)$, a colimit of $F\cdot\Phi_A$ is $FX$'

which makes perfect sense (cf. Berci's comment)


问:
Is $w$ a vector space?
Let $W$ be the set of all solutions$(X,y,i,r)$ such that $a+b=m^2$. is $w$ a vector space?
Can anybody do the whole thing for me and explain shortly every step? I have to do this kind of lots of question today.</p>

答:
It is not a vector space. Intuitively, since in a vector space everything happens linearly, you should be 'scared' of the square and think of a counter example. Here is one. The vectors
$$
(4,0,0,2), (0,0,4,2)
$$
are in your space, because $4 + 0 = 2^2$ and $0 + 4 = 2^2$. But the sum of those two vectors are $(4,0,4,4)$ and $4+4 = 8 \neq 16 = 4^2$. Therefore your space $W$ is not closed under addition, hence not a vector space.

Hope that helps,

Title: Is $w$ a vector space?


问:
How do I evaluate this summation/series?
$$\sum_{i=0}^{\infty} (1+r)^{-i} = \ ?$$

I have so far:
$$\sum_{i=0}^{\infty} (1+r)^{-i} = 1+ \frac{1}{1+r}+\frac{1}{(1+r)^2}+\ ... \ \frac{1}{(1+r)^\infty}$$

But I think it's supposed to be:
$$\sum_{i=0}^{\infty} (1+r)^{-i} = \frac{1+r}{r} $$

Edit: I think the key is that I'm failing to understand how the denominators converge to r

答:
This is a geometric series if $r >0$ but an easy approach is

$$S=1+ \frac{1}{1+r}+\frac{1}{(1+r)^2}+\frac{1}{(1+r)^3}+\cdots $$

multiply by $1+r$ to give

$$(1+r)S = (1+r)+1+ \frac{1}{1+r}+\frac{1}{(1+r)^2}+\cdots$$

subtract to give

$$(1+r)S - S=1+r$$

and rearrange $$S=\frac{1+r}{r}$$


问:
Closed-form of $\int_0^1 \int_0^1 \int_0^1 x^{(y^z)} \,dz\,dy\,dx$
We know that

$$\int_0^1 \int_0^1 x^y\,dy\,dx = \ln 2.$$

Do we know a closed-form of

$$\int_0^1 \int_0^1 \int_0^1 x^{(y^z)} \,dz\,dy\,dx\,?$$

As a start we know that

$$\int_0^1 x^{(y^z)}\,dz = \frac{\operatorname{Ei}(y \ln x) - \operatorname{li}(x)}{\ln y}.$$

Here $\operatorname{Ei}$ is the exponential integral, and $\operatorname{li}$ is the logarithmic integral.

答:
I found the answer while browsing Reddit and thought you might want to hear about it. although, I can't really prove it on my own. Edit: I managed to find a way to compute it.
$$\int_0^1 \int_0^1 \int_0^1 x^{y^z}\,dx\,dy\,dz= - \int_ {1/2}^{1} \frac{ψ(x)}{x} \,  dx - \frac{1}{2} \, \ln^{2}(2). $$
This is not a coincidence and I can prove it.
$$I=\int_0^1 \int_0^1 \int_0^1 x^{y^z}\,dx\,dy\,dz = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{∞}y^{nz}\ln\left(x\right)^{n}dxdydz
= 1-\sum_{n=1}^{∞}\left(-1\right)^{n-1}\ \frac{\ln\left(n+1\right)}{n}$$
$$ \int_{0}^{1}x^{n-1}li\left(x\right)dx=-\frac{\ln\left(n+1\right)}{n} (observation)$$
$$ I = 1+\int_{0}^{1}\frac{li\left(x\right)}{x+1}dx $$
$$I(t) = 1+\int_{0}^{1}\frac{li\left(x^t\right)}{x+1}dx , I(1)=I, I(∞)=1$$
$$ I'(t)=\frac{1}{t}\int_{0}^{1}\frac{x^{t}}{x+1}dx = \frac{\ln2+ψ\left(\frac{t}{2}+1\right)-ψ\left(t+1\right)}{t}=\frac{\ln2+ψ\left(\frac{t}{2}\right)-ψ\left(t\right)}{t}+\frac{1}{t^{2}}  $$
$$ I= 1-\int_{1}^{∞}\left(\frac{\ln2+ψ\left(\frac{t}{2}\right)-ψ\left(t\right)}{t}+\frac{1}{t^{2}}\right)dt = -\int_{1}^{∞}\frac{\ln\left(2\right)+ψ\left(\frac{t}{2}\right)-ψ\left(t\right)}{t}dt  $$
$$  F\left(t\right)=\int_{1}^{t}\frac{ψ\left(x\right)}{x}dx≃\frac{1}{2}\left(ψ\left(\frac{t}{2}+1\right)^{2}-ψ\left(t+1\right)^{2}\right)  ,F(1)=0 $$
$$ -\lim _{t\to \infty }\left(ln\left(2\right)ln\left(t\right)+F\left(\frac{t}{2}\right)-F\left(t\right)\right)+F\left(\frac{1}{2}\right)-F\left(1\right) $$
$$ (\lim _{t\to \infty }\left(ψ\left(\frac{t}{2}+1\right)-ψ\left(t+1\right)\right)=-ln\left(2\right)) $$
$$ =-\lim _{t\to \infty }\left(\ln \left(2\right)\ln \:t+\frac{1}{2}\left(ψ\left(\frac{t}{2}+1\right)^2-ψ\left(t+1\right)^2\right)\right)$$
$$ =-\lim _{t\to \infty }\left(\ln \:t-\frac{1}{2}ψ\left(\frac{t}{2}+1\right)-\frac{1}{2}ψ\left(t+1\right)\right)ln\left(2\right) $$
$$ =-\lim _{t\to \infty }\left(\frac{1}{2}ln\left(\frac{t}{2}\right)+\frac{1}{2}ln\left(2\right)+\frac{1}{2}ln\left(t\right)-\frac{1}{2}ψ\left(\frac{t}{2}+1\right)-\frac{1}{2}ψ\left(t+1\right)\right)ln\left(2\right) $$
$$\lim _{x\to \infty }\left(ln\left(x\right)-ψ\left(x+1\right)\right)=0 $$
$$= -\frac{1}{2}ln\left(2\right)^2+F\left(\frac{1}{2}\right) = -\frac{1}{2}ln\left(2\right)^2-\int _{\frac{1}{2}}^1\:\frac{ψ\left(x\right)}{x}dx$$
Edit 2: The answer can also be written in terms of generalized Stieltjes constant.
$$ γ_1\left(a,b\right)=\lim _{n\to \infty }\left(-\frac{1}{2}ln\left(n\right)^2+\sum _{k=1}^n\:\frac{ln\left(k+a\right)}{k+b}\right) $$
$$ I=1+2γln\left(2\right)-\frac{1}{2}ln\left(2\right)^2-γ_1\left(1,0\right)+γ_1\left(\frac{1}{2},0\right) $$


问:
Show that two orthogonal probability measures have the following representations using Vitali covering
Problem:

$\alpha$ and $\beta$ are two mutually orthogonal probability measures in the sense that for some Borel subset $E\subset[0,1]$, $\alpha(E)=0$ and $\beta(E)=1$. If $F(x)=\alpha\{[0,x]\}$ and $G(x)=\beta\{[0,x]\}$ for $0<x\leq 1$, show that for any $\varepsilon>0$, there is finite collection of disjoint intervals $\{[a_i,b_i]\}$ s.t. $$\sum_iF(b_i)-F(a_i)<\varepsilon\text{ and }\sum_iG(b_i)-G(a_i)>1-\varepsilon$$

Attempt:

I'm using the following version of Vitali covering:

Given a Vitali covering $\mathcal{U}$ of a set $A\subset[0,1]$ and given any $\varepsilon>0$, there is a finite collection of disjoint intervals $\{E_i\}_{i=1}^{N}$ from $\mathcal{U}$ s.t. $$\mu^*\left(A\Delta\bigcup_{i=1}^{N}E_i\right)<\varepsilon$$

By Caratheodory extension, if $\mu$ is also a measure on $\sigma(\mathcal{R})$, the sigma ring generated by the semiring, $\mu^*=\mu$ on $\sigma(\mathcal{R})$. (???)

Apply this version of Vitali covering lemma to $\mathcal{R}:\{(a,b]:0\leq a<b\leq 1\}$, the semiring, then $\sigma(\mathcal{R})$ is the Borel set. Then we consider $\mu=\frac{\alpha+\beta}{2}$. By Vitali covering, there exists disjoint $\{(a_i,b_i]\}_{i=1}^{N}$ s.t.
$$\mu\left(A\Delta\bigcup_{i=1}^{N}(a_i,b_i]\right)<\frac{\varepsilon}{2}$$
which gives
$$\alpha\left(A\Delta\bigcup_{i=1}^{N}(a_i,b_i]\right)<\varepsilon\text{ and }\beta\left(A\Delta\bigcup_{i=1}^{N}(a_i,b_i]\right)<\varepsilon$$
Hence
$$\sum_{i=1}^{N}\left[F(b_i)-F(a_i)\right]=\alpha\left(\bigcup_{i=1}^{N}(a_i,b_i]\right)=\alpha\left(\bigcup_{i=1}^{N}(a_i,b_i]\setminus A\right)+\alpha(A)<\varepsilon$$
and
$$\sum_{i=1}^{N}\left[G(b_i)-G(a_i)\right]=\beta\left(\bigcup_{i=1}^{N}(a_i,b_i]\right)=\beta(A)-\alpha\left(A\setminus\bigcup_{i=1}^{N}(a_i,b_i]\right)>1-\varepsilon$$

Question:

However, I am not so sure how to extend the outer measure given in Vitali covering (which applies to Lebesgue outer measure) to the probability measure on $[0,1]$. Can I just apply Caratheodory extension like just I did, or is it wrong and are there any other ways to do that? Other methods without using Vitali covering are also welcome. Thank you.

答:
Here's an attempt at a solution:

Use the outer regularity of $\alpha$ to pick an open set $K \supset A$ such that $\alpha(K) < \alpha(A) + \epsilon/2 = \epsilon/2$. Since $K$ is open, there exists a countable collection of intervals $\{[a_i, b_i]\}_{i=1}^{\infty}$ with $K = \cup_{i=1}^{\infty}[a_i, b_i]$ and therefore there exists a finite  sub-collection $\{[a_i, b_i]\}_{i=1}^{n}$ that covers $K$:

$$ K \subset \bigcup_{i=1}^{n}[a_i, b_i].$$

Now we can choose $N \geq n$ sufficiently large so that

$$\alpha(K) < \alpha \left( \bigcup_{i=1}^{N}[a_i, b_i]  \right) = \sum_{i=1}^N F(b_i) - F(a_i) < \epsilon.$$

I don't think we even need Vitali covering here, but I could be wrong. Not quite sure how to reconcile the issue of the intervals being disjoint, which is where Vitali might come in.

Proving the inequality for $\beta$ involves a similar argument, perhaps using the inner regularity instead of $\beta$ instead of outer regularity.

Hope this helps a bit! I'm interested in discovering the answer as well, so will follow this.


答:
This is a geometric series, and can be evaluated using the formula
$$\sum_{n\ge0}x^n=\frac{1}{1-x},\qquad{|x|<1}.$$
We see that your sum is given by plugging in $x=\frac{1}{1+r}$. Using this, we must make the assumption that $r$ satisfies $|1/(r+1)|<1$. Thus we have
$$\sum_{i\ge0}(1+r)^{-i}=\frac{1}{1-\tfrac{1}{1+r}}=\frac{r+1}{r}.$$

Title: How do I evaluate this summation/series?

Body:
$$\sum_{i=0}^{\infty} (1+r)^{-i} = \ ?$$

I have so far:
$$\sum_{i=0}^{\infty} (1+r)^{-i} = 1+ \frac{1}{1+r}+\frac{1}{(1+r)^2}+\ ... \ \frac{1}{(1+r)^\infty}$$

But I think it's supposed to be:
$$\sum_{i=0}^{\infty} (1+r)^{-i} = \frac{1+r}{r} $$

Edit: I think the key is that I'm failing to understand how the denominators converge to r


问:
How to prove that $B_{d_{H}}(x, \varepsilon) = B_{d}(x, \varepsilon) \cap H$.
Let (X, d) be a metric space and H$\subset$X. ¿How can I prove the question above?

I’ve tried proving that the first set is included in the second one, but I’m not sure if I can say that $\{h\in H / d(x,h) < \varepsilon\} \subset \{a \in X / d(x,a) < \varepsilon \} \cap H$ just by stating that every $h \in H$ is also in X since H is a subset of X.

Also, I don’t know how to demonstrate the second inclusion to show that both sets are equal.

答: Let me assume $\epsilon$ is $1$ and fix an $x \in X$. By definition $B_d(x, 1)\,\bigcap\,H = \{v \in X\,\vert\,d(v, x) < 1\}\,\bigcap\,H = \{v \in H\,\vert\,d(v, x) < 1\}$. Here $d_H$ is the relative metric in the subset $H$ and hence $B_{d_H}(x, 1) = \{v \in H\,\vert\,d(v, x) < 1\}$ because $d_H$ can only be applied to elements in $H$. Now you can replace $\epsilon$ by any positive real value.


问:
Prove that every embedded submanifold is naturally a manifold
it's my first question, I know it's easy but I just dont know how to write this demonstration, any help is welcome, thanks!</p>

答:
$\textbf{Hint}$

In order to find charts for the submanifold try to use the embedding and the charts of the ambient manifold...