问:
How would I go about solving this Integral?
I have just come out of a test where this was one of the questions, I have given it more thought at home and I still don't really see what rule I should use to solve this.

The question reads roughly as:

Given $f:[0, 1]\to (-6,6)$ Calculate the following integral if $f(0)=\sqrt{35}$ and $f(1)=\sqrt{27}$

$$\int_0^1\frac{(f(x))^3f'(x)}{\sqrt{36-(f(x))^2}} \, dx$$

I am not sure about the conditions, so those might be incorrect, but the integral is this one for sure. I was only able to get to the realization that the integral was equivalent to: $\int_0^1\frac{\frac{d}{dx}(f(x))^4}{\frac{d}{dx}(\sin^{-1}(\frac{f(x)}{6}))} \, dx$

答:
Using the hint in comments, substitute $u=f^2,\frac{du}{dx}=2ff'$:
$$I=\frac12\int_{35}^{27}\frac u{\sqrt{36-u}}\,du=\frac12\cdot\frac{-380}3=-\frac{190}3$$


问:
what are all the continuous functions $f(x)$ that is has a domain $x>0$ and $f(ab)=\frac{f(a^2)+f(b^2)}{2}$
what are all the continuous functions $f(x)$ that is has a domain $x>0$ and $f(ab)=\frac{f(a^2)+f(b^2)}{2}$.

this question was changed because the answer to the original problem what are all the continuous functions $f(x)$ that is has a domain $x>1$ and $f(ab)=f(a^2)+f(b^2)$ was trivial and User$8128$ pointed out

I think it would slightly more interesting to consider $f(ab) = \frac{f(a^2) +f(b^2)}{2}$ That way, there is a formal similarity to the inequality $ab \le \frac{a^2 + b^2}2$

答: Write $g(x):=f(e^x)$ to restate the problem as $g((x+y)/2)=(g(x)+g(y))/2$, with $g$ continuous on $\Bbb R$. Finish with this question, which proves the suitable $g$ are precisely the linear $g$, in accord with @ProfessorVector's characterization of the suitable $f$.

ContentLicense: CC BY-SA 4.0


问:
What is the maximun size of an inverse matrix that can be calculated using the Cofactor Method?
During the third lecture of 18.02SC on MIT OpenCourseWare at 43:45 it is said that the cofactor method for finding matrix inverses is not suitable for large matrices. What is the limit size of a matrix used by algorithms using the cofecator method to calculate matrix inverses. The link of the lecture is the following: https://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-3-matrices/.</p>

答:
There's no hard limit, it's just that the number of operations required grows more quickly than other methods.

Both determinants and inverses can be computed by row-reducing a matrix. Say that on one particular computer, this approach takes $1$ second for a $1000 \times 1000$ matrix. (It's enough to put the matrix in row-echelon form for a determinant, but for the inverse, we have to put a $1000 \times 2000$ matrix in reduced row-echelon form. So let's say the inverse takes several seconds, in the interests of fairness.)

The cofactor method for computing the inverse of such a matrix requires finding $1000^2$ determinants of $999\times999$ matrices, which will each take nearly the same time. You can still use the cofactor method for matrices of this size, but on the same computer, this will take about a million seconds, or over $11$ days. You would have to be very patient.


问:
Calculating the flux of the curl

Consider the sphere with radius $\sqrt{2}$ and centre the origin.  Let $S'$ be the portion of the sphere that is above the curve $C$
(lies in the region $z \geq 1$) and has $C$ as a boundary. Evaluate the flux of $\nabla \times F$ through $S_0$. Specify which orientation you are using for $S'$.

$$F = (z-y,0,y)$$

(curve $x^2+y^2=1$ lying in the plane $z=1$).

So my thoughts are can we use the Divergence Theorem? If so when we use the $\text{curl} = \langle 1,1,1\rangle$?

Then when we apply the Divergence Theorem we can just say that the flux of this would be $0$? Or is this incorrect?

答:
Method1:

Parametrize the surface as $(\sqrt2 \cos\theta \sin \phi, \sqrt2 \sin\theta \sin \phi,\sqrt2 \cos \phi)$

Outward unit normal vector $\hat{n} = (\cos\theta \sin \phi, \sin\theta \sin \phi,\cos \phi)$

At $z = 1, \phi = \frac{\pi}{4}$

So use $0 \leq \phi \leq \frac{\pi}{4}, 0 \leq \theta \leq 2\pi$ to set up your double integral.

$\displaystyle \iint_R (\nabla \times \vec{F}) \cdot \hat{n} \ (\sqrt2)^2 \ d\phi \ d\theta$

Or Method 2

Take boundary C which is a circle of radius $1$ and parametrize as $r(t) = (\cos t, \sin t, 1), 0 \leq t \leq 2\pi$.

$\displaystyle \int_C \vec{F} \cdot r'(t) \ dt$

Please make sure the orientation matches up for both integral.

Title: Calculating the flux of the curl


问:
Solve the equation: $\sin 3x=2\cos^3x$
Solve the equation  :

$$\sin 3x=2\cos^3x$$

my try :

$\sin 3x=3\sin x-4\sin^3x$

$\cos^2x=1-\sin^2x$

so:

$$3\sin x-4\sin^3x=2((1-\sin^2x)(\cos x))$$

then ?

答:
1st of all you need to learn the formula Sin 3x and remember it so that whenever you see such a question you will choose the right way.
1st learn it . This may help you to learn this
https://youtu.be/He4JXYBwTj4

Come to our answer

sin 3x = 2 cos³ x

3sinx-4sin³x=2cos³x

(here look at the expression should you expand it in terms of cubic terms, obviously no it will mess up as in your case, it will be like doing so much thing for getting a small result. So here we will rather than expanding it we will divide cos³x on rhs. )

3tanxsec²x-4tan³x=2

Let tan x = t

So,

3 t (1+t²)-4t³=2
Then solve it and get

Tan x = 1 and -2

Hope it helps.

ContentLicense: CC BY-SA 3.0

Title: Solve the equation: $\sin 3x=2\cos^3x$


问:
The degree of an irreducible representation of a finite solvable group
I know that the degree of an irreducible representation of a finite group $G$ over a field of characteristic $0$ divides the order of $G$. I also know that such a result is false when the characteristic $p$ of the field is positive (even if $p$ does not divide the order of $G$). However, such examples seem all to be non-solvable and actually searching online I've found clues of the fact that for a solvable group it is indeed possible that the degree of an irreducible representation (over a field of positive characteristic) divides the order of the group itself. Since I cannot find anything better than clues, I'm asking: is this true? Where is it possible to find a proof of this fact?

答:
It is true (with the caveats given in the other answer):
Let $G$ be a finite solvable group, and $p$ a prime dividing the order of $G$. Then the degree of every absolutely irreducible representation (that is, the field sufficiently large) of $G$ in characteristic $p$ divides the order of $G$.

The only proof for this that I am aware of, is a consequence of the Fong-Swan theorem, which states that for solvable groups actually the representation itself must come from reducing a representation in characteristic zero (and then using the result for characteristic zero). This theorem is (not the original reference, but probably the easiest accessible place) Theorem 38 on p. 135 of Serre, Linear Representations of Finite Groups.


问:
Calculating Win Probability given max possible points
I am making an app for a series of game nights my friends and I are having, and wanted to add a win probability chart.   The rules are rather simple.   Each person hosts one game night and points are awarded based on where you finish (1st: 4pts, 2nd: 3pts, 3rd: 2pts, 4th: 1pt).   Given that there are four rounds, what would be the best way to track a persons win probability for the whole tournament based on points they have scored thus far.   I dont have a background in statistics and would appreciate any help you can give, thank you!</p>

答:
This is the problem of points and is generally considered the founding problem of probability theory.  There's a discussion of various methods of solution here.  With so few possibilities, I would recommend the approach of Fermat.  Simply list all possible outcomes of the future games and figure out who wins in each.

Calculating the probability of each outcome presents a problem.  You say that the four games are different, but that you want to use the results of the past games to estimate the probabilities that the players will win future games.  If a player is good at one game, does it follow that he's good at another?  I'm dubious.  If they were playing the same game repeatedly, it would make more sense.  There's a discussion of this here.

I'd be inclined to assume each player is equally to win each future game (or to place second, third or last.)


问:
Is there any Simplification to the sum
Can we simplify the following sum $$\sum_{k=0}^{10}{20\choose k}(1/4)^k (3/4)^{20-k}$$
MY TRY: I tried to reduce this expression using the fact that $20\choose k$ = $20\choose20-k$.

Going by the expansion of $(1/4+3/4)^{20}$, we get

$$(1/4+3/4)^{20}=1= \sum_{k=0}^{20}{{20\choose k}(1/4)^k (3/4)^{20-k}}$$  $$1=
 \Bigr(\sum_{k=0}^{9}{{20\choose k}\bigr((1/4)^k (3/4)^{20-k}\bigl)+\bigr((1/4)^{20-k} (3/4)^{k}\bigl)}\Bigr)+ {20\choose 10}(1/4)^{10} (3/4)^{10}$$

I am stuck over here. How to simplify this from here?

Any hint would be a great help!

答: I would think that an obvious point would be to combine $\left(\frac{1}{4}\right)^k\left(\frac{3}{4}\right)^{20-k}$ as $\frac{3^{20- k}}{4^{20}}$.

Title: Is there any Simplification to the sum


问:
Given two points in three dimensions, how can one find the direction of motion and angle of descent of a projectile?
Say you shoot a basketball and you have the data for the ball's x, y, and z coordinates (where z is height and peaks at around z=16) through its time in the air. That's the data that I have. I want to take the ball's x,y,z coordinates when a certain z is reached on its descent, and compute the direction of motion (in terms of x and y) and angle of descent (at what angle is the ball falling at this point? like how steep is the fall) using data from that point and the previous point.

In other words, I have two points taken from the descent of a projectile: $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$, and I want to compute the descent angle and the direction of motion.

I found the following formula for descent angle: $\theta=cos^{-1}(\frac{(x_1x_2+y_1y_2+z_1z_2}{\sqrt{x_1+y_1+z_1}\sqrt{x_1+y_2+z_1}})$. However, I'm getting outputs of less than 1 degree while I'd expect the descent to be pretty steep so I'm not sure if this formula is appropriate for this problem.

For direction of motion, I'm thinking maybe I should just calculate the slope of the points $(x_1,y_1)$ and $(x_2,y_2)$ such that $\text{dir}=\frac{y_2-y_1}{x_2-x_1}$, but again I'm not 100% sure, especially because the answer would not be in degrees.

I also have data for time at these points if that is needed.

答:
You made a few errors in transcribing the standard formula
$$\arccos\left(\frac{x_1x_2+y_1y_2+z_1z_2}{\sqrt{x_1^2+y_1^2+z_1^2}\sqrt{x_2^2+y_2^2+z_2^2}}\right). $$
But your suspicion that it was not applicable is correct.
This is the formula for the angle between the vectors
$(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$,
that is, it tells you, if you were sitting at the point $(0,0,0)$
watching the projectile through a telescope,
how much you would need to turn the telescope in order to keep it pointed
at the projectile.
If the positions $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$
are not too close to $(0,0,0)$ and are observed very close in time,
then naturally you will not have to turn the telescope very much in order to track the projectile.

To find the direction of travel your projectile, a more useful vector would be the vector from one observed point to the next observed point,
$$ (x_2 - x_1, y_2 - y_1, z_2 - z_1). $$
This vector is the sum of a horizontal vector $(x_2 - x_1, y_2 - y_1, 0)$
and a vertical vector $(0,0, z_2 - z_1)$.
The vector sum can be drawn graphically as a right triangle with those two vectors as the legs and the vector $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$ as the hypotenuse.
The length of the hypotenuse is therefore
$$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} $$
and the length of the vertical leg is $z_2 - z_1$, so by simple trigonometry
the angle from straight-up vertical is
$$ \arccos\left(\frac{z_2 - z_1}
                {\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}\right) $$
and the angle from the horizontal is
$$ \arcsin\left(\frac{z_2 - z_1}
                {\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}\right). $$

But since the projectile travels in a curve rather than a straight line,
the angle of this vector is only a kind of "average" angle of travel over the arc from the first point to the second point, not the actual angle at either endpoint.
Assuming a simple parabolic arc for the path of the projectile,
the angle you get from the formulas above is the exact angle of travel at the point
$$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} + \zeta\right)$$
for some $\zeta> 0,$ that is, at some point whose horizontal coordinates are midway between the two observed points, but whose vertical position is somewhere above the line connecting the two points.
How far above the line that position will be is a function of the acceleration of gravity and of how much time passed between the observations.

If you know the acceleration of gravity and the time that passed between observations then you can work backward from the angle measured in the above formulas to the angle at the point $(x_1,y_1,z_1)$.
But a simpler method for a parabolic arc is that if you want the direction of travel at a particular time $t$, don't use the position at time $t$; instead, use the positions at time $t - \delta$ and at $t + \delta$,
because then the point at which the slope of the arc matches the slope of the vector will be the point at time $t.$

For the horizontal direction, you have the right idea about looking at the points $(x_1,y_1)$ and $(x_2,y_2)$ in the $x,y$ plane.
It's also true that the slope of the line between those points is not an angle in the usual meaning of the word "angle."
In particular, I think you would want to distinguish a trajectory that passes through $(x_1,y_1)$ first and later through $(x_2,y_2)$ from a trajectory that passes through  $(x_2,y_2)$ first and then $(x_1,y_1)$;
but the formula $(y_2 - y_1)/(x_2 - x_1)$ gives the same answer in both cases.

For the horizontal direction I suggest you consider the direction of the vector $(x_2 - x_1, y_2 - y_1)$ in the $x,y$ plane.
There is some discussion of the angular direction of this vector in
How do we really get the angle of a vector from the components?.
To summarize, the angle is essentially the result of the inverse tangent
(aka arc tangent) function.


问:
The number 3 can be written as $3$, $2+1$, $1+2$ or $1+1+1$ in four ways. In how many ways can the number $n$ be written?
Attempt
Let $x$ be any variable
$X+0=n ; X+Y=n ; X+Y+Z=n ; \dots; X+Y+Z+A+\dots=n$ (sum of n-1 terms); $1+1+1+.......+1=n$ (sum of n terms).
So total number of ways=
$$(n-1) C (1-1)+(n-1) C (2-1)+\dots+(n-1) C (n-1)$$
This equation is correct & gives correct answer if we have a value of $n$ but equation in $n$ needs to be proven equal to $2^{n-1}$.
Answer is $$2^{n-1}$$
What can be further done to solve it ?

答: Also to close the gap between your answer and the desired form, the combinatorial identity: $$ \sum_{k=0}^n \binom nk = 2^n$$ is fairly well known; the options of choosing any number of objects from a set is clearly counting the powerset of $n$ objects.


问:
Are $\mathbb Z\times \mathbb Z/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$ as groups?
Is $(\mathbb Z\times \mathbb Z)/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$,

where $4\mathbb Z\times 5 \mathbb Z$ is given as a subgroup of $\mathbb Z\times \mathbb Z$?

My feelings says that I need to find well defined $f:\mathbb Z\times \mathbb Z\to \mathbb Z$ which has a kernel $4\mathbb Z\times 5\mathbb Z$

So using first isomorphism theorem I can show the isomorphism.

I could not find the map and how can I show it in different way?

答: There shouldn't be such a map. You may want to check for elements with finite order in both groups.</p>


问:
Estimating $\int_{\mathbb{R}}|\frac{\log|x|}{1+x^4}|\ dx$
I'd like to show that $\frac{\log|x|}{1+x^4} \in L^1(\mathbb{R})$, but I have trouble with estimating
$$\int_{\mathbb{R}}\left|\frac{\log|x|}{1+x^4}\right|\ dx.$$
Writing $(1+x^4) = (x^2 - i)(x^2 + i)$ or splitting the integral into
$$\int_{\mathbb{R}}\left|\frac{\log|x|}{1+x^4}\right|\ dx = \int_{0 < |x| < 1}\frac{|\log|x||}{1+x^4}\ dx + \int_{|x| > 1}\frac{|\log|x||}{1+x^4}\ dx$$
doesn't seem to help me either. Is it true that $$\frac{|\log|x||}{1+x^4} < \frac{1}{x^2}$$
for any $|x| > 1$? If that's true, how so? What can I do with $$\int_{0 < |x| < 1}\frac{|\log|x||}{1+x^4}\ dx?$$

答:
for $|x|>1$ it is clear that $\ln|x|<x^2$ since $\ln|x|<|x-1|$ so you can use this. In your case you can use this to show that:
$$\int\limits_{|x|>1}\left|\frac{\ln|x|}{x^4+1}\right|dx<\int\limits_{|x|>1}\frac1{x^2}dx$$
now you need to try and do the other proportion of the integral where $|x|<1$


问:
Notation of Planes
I have to finde the line $L := E_0 \cap E_1$, with

$E_0 = \left\langle \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \right\rangle ^\perp$

$E_1 = \left\langle \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \right\rangle ^\perp + \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

But I am unfamiliar with the notation of $E_0, E_1$. I know the notations which can be found in this wiki article.

Does anyone know this notation, and can point me into the right direction?

答:
According to the definition, if $W\subset V$ which $V$ is a proper vector space, then $$W^\perp=\{v\in V\mid \langle v,w\rangle=0, ~\forall w\in W\}$$ Here $\langle v,w\rangle$ means the inner product of two vectors. For example, if $W=\{(0,0,z)\mid z\in\mathbb R\}\in\mathbb R^3$ be our usual $z-$ axes in $\mathbb R^3$, then we can find out that

$$W^\perp=\{(a,b,0)\mid~~a,b\in\mathbb R\}$$

Note that $W^\perp$ is a subspace of $V$.

Title: Notation of Planes


问:
$xy'' + y' = 0$, Why is this in Euler-Cauchy form?
I am having difficulty figuring out why $xy''(x) + y'(x) = 0$ is in Euler-Cauchy form. Can anyone explain this to me?  Thank you!

Body:
This is simply $$\left(x y'(x)\right)'=0$$
so $$xy'(x)=c_1$$
$$y'(x)=\frac{c_1}{x}$$
$$y(x)=c_1\log x +c_2$$


问:
What is $P[B(t)>y, \land_{0 \leq u \leq t} B(u)>0|B(0)=x]$, where $B(t)$ i standard Brownian motion?
What is $P[B(t)>y, \land_{0 \leq u \leq t} B(u)>0|B(0)=x]$, where $B(t)$ i standard Brownian motion? Here notation $\land_{0 \leq u \leq t}B(u)$ is the minimum of the process between $0$ and $t$.

Is the following attempt correct?

$P[B(t)>y, \land_{0 \leq u \leq t} B(u)>0|B(0)=x] = P[B(t)>y, \lor_{0 \leq u \leq t} B(u)<2x|B(0)=x]$,

using symmetry, where $\lor_{0 \leq u \leq t}B(u)$ is the maximum of the process. Further:

$P[B(t)>y, \lor_{0 \leq u \leq t} B(u)<2x|B(0)=x] = 1-P[B(t)>y, \lor_{0 \leq u \leq t} B(u)\geq2x|B(0)=x)$
$= 1-P[B(t)>y, |B(t)|\geq2x|B(0)=x]$,

where the last equality follows by the reflection principle. Finally, if we subtract the starting position:

$1-P[B(t)>y, |B(t)|\geq2x|B(0)=x] = 1-P[B(t)>y-x, |B(t)|\geq x]$
$= 1-P[-x<B(t)<\max(y-x, x)]$.

答:
First, subtracting the starting position, we are looking for $P(B(t)>y-x, \min_{u\leq t}B(u)>-x)$ or, by symmetry, $P(B(t)<x-y, \max_{u\leq t}B(u)<x).$ Hence,

\begin{align} P\big(B(t)<x-y, \max_{u\leq t}B(u)<x\big)=&P\big(B(t)<x-y\big)-P\big(B(t)<x-y, \max_{u\leq t}B(u)\geq x\big) \\ =& P\big(B(t)<x-y\big)-P\big(B(t)>x+y\big), \end{align}
where we used the fact that $P\big(B(t) a+b\big)$ (see, for example, Proposition 2 here).

Body:
First, subtracting the starting position, we are looking for $P(B(t)>y-x, \min_{u\leq t}B(u)>-x)$ or, by symmetry, $P(B(t)<x-y, \max_{u\leq t}B(u)<x).$ Hence,

\begin{align} P\big(B(t)<x-y, \max_{u\leq t}B(u)<x\big)=&P\big(B(t)<x-y\big)-P\big(B(t)<x-y, \max_{u\leq t}B(u)\geq x\big) \\ =& P\big(B(t)<x-y\big)-P\big(B(t)>x+y\big), \end{align}
where we used the fact that $P\big(B(t)< a-b, \max_{u\leq t}B(u)\geq a\big)=P\big(B(t)> a+b\big)$ (see, for example, Proposition 2 here).


问:
Galois group of $x^3 - 2 $ over $\mathbb Q$
I know the Galois group is $S_3$. And obviously we can swap the imaginary cube roots. I just can't figure out a convincing, "constructive" argument to show that I can swap the "real" cube root with one of the imaginary cube roots.

I know that if you have a 3-cycle and a 2-cycle operating on three elements, you get $S_3$. I have a general idea that based on the order of the group there's supposed to be at least a 3-cycle. But this doesn't feel very "constructive" to me.

I wonder if I've made myself understood in terms of what kind of argument I'd like to see?

答: I am not sure if this is too late. But the symmetry can be seen by plotting on the complex plane. The roots form the vertices of the equilateral triangle. So, you can rotate all of them and preserve symmetry. Hand-wavy but I guess it works.</p>


问:
$S$ and $T$ are idempotent linear operators in $V$ which is a vector space in $\mathbb{C}$. Prove that if $S+T$ is idempotent then $ST=TS=0$.
If $S+T$ is idempotent, then
$$(S+T)^2=S+T$$
$$\implies S^2+ST+TS+T^2=S+T$$
$$\implies S+T+ST+TS=S+T$$
$$\implies ST+TS=0$$

Now, from here how do I show that $ST=TS=0$?

答:
We have
$$
ST + TS = 0 \implies\\
S[ST + TS] = 0 \implies\\
ST + STS = 0.
$$
Similarly, we have $TS + STS = 0$. Putting these results together yields
$$
ST = -STS = TS.
$$
That is, we indeed have $ST = TS$. Since $ST + TS = 0$, conclude now that $ST = 0$.


答:
What is the kernel of the group homomorphism $\mathbb Z \times \mathbb Z \rightarrow \mathbb Z/4\mathbb Z \times \mathbb Z/5\mathbb Z$ given by

$$(x,y) \mapsto (x+4\mathbb Z, y +5\mathbb Z)?$$

Title: Are $\mathbb Z\times \mathbb Z/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$ as groups?

Body:
Is $(\mathbb Z\times \mathbb Z)/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$,

where $4\mathbb Z\times 5 \mathbb Z$ is given as a subgroup of $\mathbb Z\times \mathbb Z$?

My feelings says that I need to find well defined $f:\mathbb Z\times \mathbb Z\to \mathbb Z$ which has a kernel $4\mathbb Z\times 5\mathbb Z$

So using first isomorphism theorem I can show the isomorphism.

I could not find the map and how can I show it in different way?


问:
What is $\sum_{n \geqslant 0} x^{2^n}$?
I recently thought about the function

\begin{align*}
F(x) := \sum_{n = 0}^\infty x^{2^n} = x + x^2 + x^4 + x^8 + x^{16} + \cdots, \quad \quad x \in I = (-1,1).
\end{align*}

The series defining $F$ converges uniformly on any $\left[a, b \right] \subset I$, so $F$ is continuous on $I$.

Is there a closed-form expression for $F(x)$? Is this function somehow related to any "known" functions?

One vague idea that I got was to try to mimic the derivation of the formula for a geometric series, but doing so involves using the fact that

\begin{align*}
F(x)-F(x^2) = x
\end{align*}

and finding an inverse to the operator $\text{id} - D$ where $D\left[ F \right] (x) = F(x^2)$. (In particular it seems that one would also have to find the right space for $F$ to live in and for $\text{id}$ and $D$ to act on, together with a reasonable topology.)

答:
Not an answer, just a plot

I got a fine graph with Mathematica, below. It's so fast the convergence, that I needed to add just a few terms for any $x$ in the convergence interval $(-1,1)$.

Title: What is $\sum_{n \geqslant 0} x^{2^n}$?


问:
Is there always a "k-sparse" subsequence of a multilimit convergent sequence?
Sorry for the bad lingo. Couldn't find the right terminology.

What I mean by multilimit is:

Let $\{a_n\} \subset \mathbb{R}$, and define:
$$
L = \{p \in \mathbb{R} \quad | \quad \exists \{b_n\} \subset \{a_n\}  \quad b_n \longrightarrow p\}
$$
then L is the multilimit of $a_n$ and this I note as $a_n \longrightarrow L$

What I mean by "k-sparse" subsequence is the following:

Let $a_n$ be a real valued sequence and consider $\{b_n\} \subset \{a_n\}$ such that every term in $b_n$ is every $k$-th term in $a_n$  with $k \in \mathbb{N}$.

My question is:

If $a_n \longrightarrow L \subset \mathbb{R}$, is it true that there must be a $k$-sparse subsequence of $a_n$ such that it converges to some real value in L?

Also if you know what is the correct terminology for this let me know.

答:
It depends on the properies of $\{a_n\}$.

If $\{a_n\}$ converges to some limit $a$, then so would all subsequences of $\{a_n\}$, and thus the set $L=\{\hspace{1mm}p\hspace{1mm}|\hspace{1mm} \exists\{b_n\}\subset\{a_n\},\hspace{1mm} b_n\to p\}$, would be the singelton set $\{a\}$. In other words, all "k-sparse" subsequences would converge to $a\in L$.

However, if $\{a_n\}$ does not converge then there is not much we can say without more restrictions on $\{a_n\}$.

Example 1: let $\{a_n\}=(-1)^n$. Then $L=\{1,-1\}$, and the "2-sparse" subsequence $(-1)^{2n}$ converges to $1\in L$.

Example 2: let $\{a_n\}=sin(n)$. Then $L$ might be empty or non-empty. In either case, all "k-sparse" subsequences $\{sin(kn)\}$ diverges so they can not be in $L$.

So we cant really say much if $\{a_n\}$ does not converge. There might exists such a "k-sparse" subsequence or there might not.

Also, the notation $a_n \rightarrow L$ is a bit confusing, I would advice to use something else.


问:
Density and distributions of those numerically or analytically KNOWN solutions of Riemann $\zeta(1/2 + r i)=0?$
We know the conjecture about the Riemann hypothesis is about the nontrivial zeros are on
$$(1/2 + r i)$$
for some $r \in \mathbb{R}$ of Riemann zeta function.

My question is how much is known about the density and the distributions of those numerically or analytically KNOWN solutions of
$$\zeta(1/2 + r i)=0?$$

I found a related post but it was about 8 years ago, so maybe we have a better update?

Mean density of the nontrivial zeros of the Riemann zeta function

答:
Mathematica 8.0.1 derivation of Eric Weisstein's approximation for Gram points:

```<code>(*Start*)
(*Mathematica*)
(*The derivation of the Gram points approximation by Weisstein in \
Mathworld*)
Clear[x, n, a, g, t];
Series[RiemannSiegelTheta[x], {x, Infinity, 12}]
a = Normal[Series[RiemannSiegelTheta[x], {x, Infinity, 0}]]
g = FullSimplify[(x /. Solve[a == (n)*Pi, x])[[1]]]
n = Range[42] - 2;
t = N[g, 20]
Zeta[1/2 + I*t]
(*End*)
</code>```

9.6769067871658668471,

17.847836512849620314,

23.171660819240722718,

27.671198036307304064,

31.718791394674873194,

35.467863110275089697,...

Modified Mathematica 8.0.1 derivation of Eric Weisstein's approximation giving Franca-LeClair points:

```<code>(*Start*)
(*Mathematica*)
(*Analogous to the derivation of the Gram points approximation by \
Weisstein in Mathworld*)
Clear[x, n, a, g, t];
Series[RiemannSiegelTheta[x], {x, Infinity, 12}]
a = Normal[Series[RiemannSiegelTheta[x], {x, Infinity, 0}]]
g = FullSimplify[(x /. Solve[a == (n + 1/2)*Pi, x])[[1]]]
n = Range[42] - 2;
t = N[g, 20]
Zeta[1/2 + I*t]
(*End*)
</code>```

14.521346953065628168,

20.655740355699557203,

25.492675432264310733,

29.739411632309551244,

33.624531888500487851,

37.257370086972976394,...

The basic difficulty in getting an accurate asymptotic for the Riemann zeta zeros is that the Riemann-Siegel theta function is not invertible.
User reuns pointed out to me that the exact asymptotic for the Riemann zeta zeros has been known for about 120 years and the exact asymptotic is the functional inverse of the Riemann-Siegel theta function, according to the French Wikipedia.

Body:
Mathematica 8.0.1 derivation of Eric Weisstein's approximation for Gram points:

(*Start*)
(*Mathematica*)
(*The derivation of the Gram points approximation by Weisstein in \
Mathworld*)
Clear[x, n, a, g, t];
Series[RiemannSiegelTheta[x], {x, Infinity, 12}]
a = Normal[Series[RiemannSiegelTheta[x], {x, Infinity, 0}]]
g = FullSimplify[(x /. Solve[a == (n)*Pi, x])[[1]]]
n = Range[42] - 2;
t = N[g, 20]
Zeta[1/2 + I*t]
(*End*)

9.6769067871658668471,

17.847836512849620314,

23.171660819240722718,

27.671198036307304064,

31.718791394674873194,

35.467863110275089697,...

Modified Mathematica 8.0.1 derivation of Eric Weisstein's approximation giving Franca-LeClair points:

(*Start*)
(*Mathematica*)
(*Analogous to the derivation of the Gram points approximation by \
Weisstein in Mathworld*)
Clear[x, n, a, g, t];
Series[RiemannSiegelTheta[x], {x, Infinity, 12}]
a = Normal[Series[RiemannSiegelTheta[x], {x, Infinity, 0}]]
g = FullSimplify[(x /. Solve[a == (n + 1/2)*Pi, x])[[1]]]
n = Range[42] - 2;
t = N[g, 20]
Zeta[1/2 + I*t]
(*End*)

14.521346953065628168,

20.655740355699557203,

25.492675432264310733,

29.739411632309551244,

33.624531888500487851,

37.257370086972976394,...

The basic difficulty in getting an accurate asymptotic for the Riemann zeta zeros is that the Riemann-Siegel theta function is not invertible.
User reuns pointed out to me that the exact asymptotic for the Riemann zeta zeros has been known for about 120 years and the exact asymptotic is the functional inverse of the Riemann-Siegel theta function, according to the French Wikipedia.


答:
Consider the map
$$\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}: (x,y) \mapsto (x+ 4 \mathbb{Z}, y + 5\mathbb{Z})$$
This map is surjective and has  kernel  $\mathbb{4}\mathbb{Z}\times 5\mathbb{Z}$, so by the first isomorphism theorem,
$$(\mathbb{Z}\times \mathbb{Z})/(4\mathbb{Z}\times 5 \mathbb{Z})\cong \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}\cong \mathbb{Z}/20\mathbb{Z}$$

The latter group contains $20$ elements and is finite, so the group you are interested in is finite. Hence, it cannot be isomorphic to $\mathbb{Z}$.


问:
A Question about Doctoral Theses in Mathematics
This is most definitely a soft question, which I'm sure may get some negative attention, and perhaps even be voted closed. However, I genuinely would like to generate answers on this matter as it concerns anyone that decides to get a doctorate in mathematics. (Since I believe this is the best forum to do so, here goes...)

Question: Does a doctoral thesis in mathematics have to contain more than an abstract, a proposition and proof of a new and exciting result?

I strongly believe that brevity is beauty to mathematicians and see no problem with a thesis that contains a lengthy abstract (~350 words) to generate enthusiasm and then jumps right into the heart of a novel and far-reaching result. The proof provided is extremely condensed in the spirit of Zagier, leaving almost all details that can be found elsewhere to be found elsewhere, in the works cited. If the result has merit, I see no reason why it can't be submitted or published in such a manner. To support my stance, I offer the following condensed theses:

David Rector, "An Unstable Adams Spectral Sequence", MIT (1966), 9 pgs.

Burt Totaro, "Milnor K-Theory is the simplest part of algebraic K-theory", Berkeley (1989), 12 pgs.

Herman Buvik, A New Proof of Torelli's Theorem, NYU (1962), 12 pgs.

Eva Kallin "A non-local function algebra" Berkeley (1963), 13 pgs.

Edmund Landau "Neuer Beweis der Gleichung $\sum_{k = 1}^{\infty} \frac{\mu(k)}{k} = 0$", Berlin University (1899), 14 pgs.

Barry Mazur, "On Embeddings of Spheres", Princeton (1959), 26 pgs.

John F. Nash "Non-Cooperative Games", Princeton (1950), 27 pgs.

Kevin Walker, "An Extension of Casson's Invariant to Rational Homology Spheres", Berkeley (1989). 29 pgs.

Although I'd appreciate to hear from anyone worth giving his two cents, I'm specifically eager to hear from those senior members of the community.

Thanks!

答:
I'm not an academic, but I got a PhD in (applied) maths over 25 years ago and I do regard myself as a mathematician (among other things).

I've always thought that the main point of a PhD is to prove that the author can do original research.  That was certainly why I was always interested in hiring people with PhDs when I used to work in the banking industry, because I wanted to hire creative people who could have their own new ideas.  So my view is that in principle, a short PhD which contains original research would be absolutely fine :-).

However, I think the problem with a short PhD is a practical one.  The university that awards the degree needs to be certain that the author is responsible for the original research in their PhD.  The less there is in the thesis, the less there is to discuss and test in the viva, so establishing that the person is the true source of the exciting new result is much harder.


问:
Proving the Leibniz Integral Rule
I've been wondering if there is a fairly simple proof or derivation for the following (called the Leibniz Rule):

$$ \frac{d}{dt} \int_{a(t)}^{b(t)}f(x,t)dx = \int_{a(t)}^{b(t)}\frac{\partial{f}}{\partial{t}}dx+ f(b(t),t)\frac{d}{dt}b(t)-f(a(t),t)\frac{d}{dt}a(t)$$

What I tried doing is integrating by parts the leftmost integral, and then applying the time derivative to what it's equal to. However, it seems the other side is very far from what the result above. A small hint would suffice as I like trying to prove things on my own.

答:
HINT: first try to see what is $\partial _y\int_{a}^y f(x,t)\mathop{}\!d x$ and $\partial _t \int_{a}^y f(x,t)\mathop{}\!d x$, the first case follows from the fundamental theorem of calculus, the latter from the continuity of $\partial_t f$ and the definition of partial derivative.

Then for $F(y,t):=\int_{a}^y f(x,t)\mathop{}\!d x$ you will get an expression for $\nabla F(y,t)$, then setting $\varphi (t):=(g(t),t)$ for some differentiable $g$ you can compute $(F\circ \varphi )'(t)$ using the chain rule.

Finally the general case follows from the decomposition
$$
\int_{h(t)}^{g(t)}f(x,t)\mathop{}\!d x=\int_{a}^{g(t)}f(x,t)\mathop{}\!d x-\int_{a}^{h(t)}f(x,t)\mathop{}\!d x
$$

We will show that $\partial _t \int_{a}^{y}f(x,t)\mathop{}\!d x=\int_{a}^y \partial _t f(x,t)\mathop{}\!d x$ when $\partial _t f$ is continuous. Note that, as the partial derivative in $t$ of $f$ is continuous then for fixed $t_0$ the function defined by

$$
q_{t_0}(x,h):=\begin{cases}
\frac{f(x,t_0+h)-f(x,t_0)}{h},&h\neq 0\\
\partial _t f(x,t_0),& h=0
\end{cases}
$$

is continuous, therefore it is uniformly continuous in the compact set $[a,y]\times [-r ,r ]$ for any chosen $r >0$, so for every $\epsilon >0$ there is some $\delta \in(0,r)$ such that $|q_{t_0}(x,h)-\partial _t f(x,t_0)|<\epsilon $ for all $(x,h)\in[a,y]\times (-\delta ,\delta )$, therefore

$$
\left|\int_{a}^y (q_{t_0}(x,h)-\partial _t f(x,t_0))\mathop{}\!d x\right|\leqslant \int_{a}^y |q_{t_0}(x,h)-\partial _t f(x,t_0)|\mathop{}\!d x<(y-a)\epsilon
$$

what shows that

$$
\partial _t \int_{a}^y f(x,t_0)\mathop{}\!d x=\lim_{h\to 0}\int_{a}^y q_{t_0}(x,h)\mathop{}\!d x=\int_{a}^y \partial _t f(x,t_0)\mathop{}\!d x
$$

Title: Proving the Leibniz Integral Rule


答:
No, there is no closed-form expression for this sum, in terms of commonly-used special functions.

As a function of the complex variable $z$, the function
$$
F(z) := \sum_{n = 0}^\infty z^{2^n} = z + z^2 + z^4 + z^8 + z^{16} + \cdots, \quad \quad |z|<1
$$
has "natural boundary" the whole circle $|z|=1$.  None of the well-known functions [including the elementary functions and the standard special functions] do this.

Title: What is $\sum_{n \geqslant 0} x^{2^n}$?

Body:
I recently thought about the function

\begin{align*}
F(x) := \sum_{n = 0}^\infty x^{2^n} = x + x^2 + x^4 + x^8 + x^{16} + \cdots, \quad \quad x \in I = (-1,1).
\end{align*}

The series defining $F$ converges uniformly on any $\left[a, b \right] \subset I$, so $F$ is continuous on $I$.

Is there a closed-form expression for $F(x)$? Is this function somehow related to any "known" functions?

One vague idea that I got was to try to mimic the derivation of the formula for a geometric series, but doing so involves using the fact that

\begin{align*}
F(x)-F(x^2) = x
\end{align*}

and finding an inverse to the operator $\text{id} - D$ where $D\left[ F \right] (x) = F(x^2)$. (In particular it seems that one would also have to find the right space for $F$ to live in and for $\text{id}$ and $D$ to act on, together with a reasonable topology.)


问:
Positive operators given by kernels
I am wondering about a proof or reference to the following claim by Getoor;
Let $(E,\mathscr E)$ and $(F,\mathscr F)$ be two measurable spaces. A (positive) {kernel} from $(F,\mathscr F)$ to $(E,\mathscr E)$ is a map
$$k: F\times \mathscr E \to [0,\infty],\quad (x,E)\mapsto k(x,E)$$
such that $x\mapsto K(x,A)$ is $\mathscr F$ measurable for each fixed $A\in \mathscr E$ and $A\mapsto k(x,A)$ is a measure on $(E,\mathscr E)$ for each fixed $x\in F$. If $k$ is a bounded kernel from $(F,\mathscr F)$ to $(E,\mathscr E)$, then
$$K: \mathcal{B}(\mathscr E) \to \mathcal B(\mathscr F),\quad K f(x)=K(x,f)=\int_E f(y) k(x,dy)$$
defines a bounded linear map (from bounded measurable to bounded measurable functions) with
$$ (f_n)_{n\in \mathbb{N}} \subseteq \mathcal{B}(\mathscr E)^+ \uparrow f \in \mathcal{B}(\mathscr E) \Rightarrow K f_n \uparrow K f.$$
Claim: Conversely every bounded linear positive map $ \mathcal{B}(\mathscr E) \to \mathcal B(\mathscr F)$ that satisfies the above condition is given by a kernel

How do you go about showing something like this? Is it some Banach algebra characterisation theorem? We can define the composition of kernels as
If $k$ is a kernel from $(F,\mathscr F)$ to $(E,\mathscr E)$ and $l$ is a kernel from $(G,\mathscr G)$ to $(F,\mathscr F)$, then the composition of $k$ and $l$ is a kernel $l\star k$ from $(G,\mathscr G)$ to $(E,\mathscr E)$ defined by
$$l\star k (x,A)=\int_F k(y,A) l(x,dy)$$
for $x\in G$ and $A \in \mathscr E$.
On the bounded kernels with the sup norm we get a submultiplicative norm. I'm guessing this is a Banach algebra with
$$L\star K f(x)=\int_F f(y) l\star k(x, dy)=\int_F f(y) \int_F k(z,dy)l(x,dz)$$
But I'm not sure this is the right way to define the composition operator

答:
For $x\in F$ and $A\in\mathscr{E}$ let $k(x,A)=K1_A(x)$. Since $K1_A$ is measurable by assumption, it suffices to check that $A\mapsto k(x,A)$ is a measure for every $x\in F$. For that purpose let $(A_n)$ be a disjoint sequence in $\mathscr{E}$ and $f_n=1_{\bigcup_{k=1}^n A_k}=\sum_{k=1}^n 1_{A_k}$. Then $f_n\nearrow 1_{\bigcup_{k=1}^\infty A_k}$, hence
$$
\sum_{k=1}^n k(x,A_k)=Kf_n(x)\nearrow K1_{\bigcup_{k=1}^\infty A_k}(x)=k\left(x,\bigcup_{k=1}^\infty A_k\right).
$$
Note that the space you constructed is not a Banach algebra because you cannot multiply arbitrary elements. In any case, even if $(E,\mathscr{E})=(F,\mathscr{F})$, I do not see how this result should follow from an abstract representation theorem for Banach algebras.

Title: Positive operators given by kernels


答:
No.

Since $$G:=\Bbb Z\times \Bbb Z\cong\langle a,b\mid ab=ba\rangle,$$

we can write the subgroup $H:=4\Bbb Z\times 5\Bbb Z$ as the normal subgroup $$\langle \langle a^4,b^5\rangle \rangle, $$ so that

$$\begin{align}
G/H &\cong \langle a,b\mid a^4, b^5, ab=ba\rangle\\
&\cong \langle a,b,c\mid a^4, b^5, ab=ba, a=cb^{-1}\rangle \\
&\cong \langle b,c\mid (cb^{-1})^4, b^5, c=b(cb^{-1})\rangle\\
&\cong\langle b,c\mid c^4=b^4, b^5, cb=bc\rangle\\
&\cong\langle b,c\mid c^4=b^{-1}, b^5, cb=bc\rangle \\
&\cong \langle b,c\mid b=c^{-4}, (c^{-4})^5, c^{-3}=c^{-3}\rangle \\
&\cong\langle c,d\mid d=c^{-1}, d^{20}\rangle \\
&\cong\langle d\mid d^{20}\rangle\\
&\cong\Bbb Z/ 20\Bbb Z\\
&\not\cong\Bbb Z.
\end{align}$$


问:
How would I go about solving this Integral?
I have just come out of a test where this was one of the questions, I have given it more thought at home and I still don't really see what rule I should use to solve this.

The question reads roughly as:

Given $f:[0, 1]\to (-6,6)$ Calculate the following integral if $f(0)=\sqrt{35}$ and $f(1)=\sqrt{27}$

$$\int_0^1\frac{(f(x))^3f'(x)}{\sqrt{36-(f(x))^2}} \, dx$$

I am not sure about the conditions, so those might be incorrect, but the integral is this one for sure. I was only able to get to the realization that the integral was equivalent to: $\int_0^1\frac{\frac{d}{dx}(f(x))^4}{\frac{d}{dx}(\sin^{-1}(\frac{f(x)}{6}))} \, dx$

答:
When you see $f'(x)\,dx,$ if you don't immediately think of $u=f(x)$ and $du = f'(x)\,dx,$ then you may have missed part of the idea of how substitutions work. This substitution yields
$$
\int_{f(0)}^{f(1)} \frac{u^3}{\sqrt{36-u^2}} \, du = \int_{\sqrt{35}}^{\sqrt{27}} \frac{u^3}{\sqrt{36-u^2}} \, du.
$$
Then if you write this as $\displaystyle\int_{\sqrt{35}}^{\sqrt{27}} \frac{u^2}{\sqrt{36-u^2}}\Big( u\, du\Big)$ then $\text{“}\Big(u\, du \Big)\text{”}$ suggests either $v=u^2$ or $v=36-u^2,$ and you can see why the latter brings about more simplification than the former. You get
$$
\int_1^9 \frac{36-v^2}{\sqrt v} \Big(\tfrac 1 2 \, dv\Big) = \frac 1 2\int_1^9 \left( 36 v^{-1/2} - v^{3/2} \right) \, dv
$$
and so on.


问:
Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$
I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.

The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$

How am I supposed to solve this problem?

答:
Denote $y = \tan x \cot 2x = \frac{1-\tan^2x}2$ and express $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$ as a system of equations in $x,y$

$$\tan x+\frac{2\tan x}{1-\tan^2x}=\left(1+\frac1y \right)\tan x=\frac{2}{\sqrt{3}}
$$
Then, eliminate $\tan x$ to get
$$\frac3{y^3} -\frac{13}{y}-6=0$$
which is a depressed cubic equation in $\frac1y$, yielding

$$\tan x \cot 2x=y= \left( \frac{2\sqrt{13}}3 \cos \left( \frac13\cos^{-1} \frac{27}{13\sqrt{13}}\right) \right)^{-1}
$$


问:
an identity related to Ricci tensor

If $$R_{ij,k}=2\lambda_kR_{ij}+\lambda_iR_{kj}+\lambda_jR_{ik}$$ where
$\lambda_i$ is a covariant vector , $R_{ij}$ is Ricci tensor , show
that $$\lambda_k=\frac{\partial}{\partial x^k}\log\sqrt{R}$$

Multiplying $R_{ij,k}=2\lambda_kR_{ij}+\lambda_iR_{kj}+\lambda_jR_{ik}$ both sides by $g^{ik}$ $$(g^{ik}R_{ij})_{,k}=2g^{ik}\lambda_kR_{ij}+g^{ik}\lambda_iR_{kj}+g^{ik}\lambda_jR_{ik}$$ $$\implies R^k_{j,k}=2\lambda_iR_{ij}+\lambda_kR_{kj}+\lambda_jR$$ By contracted Bianchi identity $$\frac{1}{2}R_{,j}=2\lambda_iR_{ij}+\lambda_kR_{kj}+\lambda_jR$$ Afterwards I have no idea how to proceed . Any help is appreciated .

答: My suggestion is to fold the initial equation also by $g^{ij}$ and use symmetry $R_{kj}=R_{jk}$. You will get the second equation $R_{,k}=2\lambda_iR_{k}^{i}+2\lambda_kR$. Then multiply the first equation that you got by 2 and extract from the second one - you will get $\lambda_iR_{k}^{i}=0$ and, consequently, from the second equation: $R_{,k}=2\lambda_kR$.

Title: an identity related to Ricci tensor


问:
How to find how much steps up and to the side by knowing angle and steps
Let's say you know how long a line is and where it's pointing.

By knowing that how do you find how much steps a line extends downward and to the side by knowing the line length and angle?

For example: an line that is 4 feet long and extends at an 35 degree angle.
How far does the line extend out to the right and up?

答:
Imagine a circle centered at $(0,0)$ with radius $1$.  Now imagine all the points of the circle labeled as $(x,y)$ and they all satisfy the equation $x^2 + y^2 = 1^2$.

Now imagine that for every point $(x_1, y_1)$ draw the line from $(0,0)$ to $(x_1,y_1)$.  That point will have a specific angle $\theta_1$.  ANd vice versa.  If you imagine an angle $\theta_2$, line for $(0,0)$ at that angle will hit the circle at exactly on point $(x_2, y_2)$.

So because there is a one-to-one correspondence there is function between angles and points defined like that.

There is no way to use algebra to figure it out but we do have a function and we make a list of all those values and we used to right them down in books but now we program them into caluculators and program languages.

There are two functions.  If you take and angle $\theta$ and imagin a line at that angle from $(0,0)$ at that angle, and extend it a distance of $1$ it will hit a circle at $(x_\theta, y_\theta)$.  We call $x_\theta = \cos (\theta)$ and wee call $y_\theta = \sin (\theta)$.

So you want to find $4\times \cos 35^\circ$ is how for it goes over, and $4\times \sin 35^\circ$ is how far it goes up.

If you punch into a calculator $\cos 35$ you get $0.81915204428899178968448838591684$ and if you punch in $\sin 35$ you get $0.57357643635104609610803191282616$.  (Notice that if you square and add them they add up to $1$.  Thats because they are the $x,y$ values of a point on a circle).

So you want $4\times 0.819= 3.276$ and $4\times 0.573=2.292$.

Notice that $3.276^2 + 2.292^2\approx 16_{\text{with small rounding error}} = 4^2$ so it satisfies Pythogorean Theorem.

The branch of mathematics called Trigonometry is entirely about these functions.

A few things to note:

$\sin^2 x + \cos^2 x = 1$.  Always.

And two.  If you have a point $(x,y)$ on a sheet of paper and your rotate it by $\theta$ angles and you want to know where $(x,y)$ end up.... well, that can be solved by figuring out what ange $\alpha$ has $\sin \alpha =y$ and $\cos \alpha = x$.  Figuring out what the distance from $(x,y)$ to $(0,0)$ is $r=\sqrt{x^2 + y^2}$ and figuring the new angle is $\alpha + \theta$ and the new point is $(r\cos(\alpha + \theta), r\sin(\alpha + \theta))$.

Trigonometry is very useful for these types of problems.


问:
Can't understand Mean Absolute Error (MAE) formulation
I'm reading Taylor Cross Entropy Loss paper and came across the formulation of Mean Absolute Error (MAE), which is described as following:

$$
\mathcal{L}_{MAE}(f({\bf x}), y) = \Vert{e_y - f({\bf x})}\Vert_{1} = 2 - 2f_y({\bf x})
$$

As mentioned in the linked paper's section 3.1, $e_y$ is one hot encoded vector having same dimension as $f({\bf x})$.

What I don't understand is that from where does the constant 2 come from in above formulation? I'm referring to this formulation if MAE on Wikipedia.

Any hint/reference would be appreciated. Thanks.

Edit

As requested in the comment, here is the context of the above formulation.

The linked paper talks about training a deep neural network for k class classification. ${\bf x}$ is a feature vector (e.g. image of cat) and $y$ is ground truth label.

The neural network is represented as an unknown complex function $f$.

$e_y$ is one hot encoded vectors of ground truths. For example, if k=2 (cat and dog), then $e_y = [0, 1]$ for cat and $e_y = [1, 0]$ for dog.

$f({\bf x})$ is what predicted by neural network. It can be probabilities for each class. For example, a well trained neural network will output $f({\bf x}) = [0.05, 0.95]$ for an image of cat.

The mean absolute difference between ground truth labels and the predicted outputs for all images is what I refer to MAE.

答:
Suppose both $e_y$ and $f(x)$ are vectors where a single element has value $1$ and all other elements have value $0$. Therefore, $\Vert{e_y - f({\bf x})}\Vert_{1}=0$ if the position of the $1$ is the same in both vectors, and $\Vert{e_y - f({\bf x})}\Vert_{1}=2$ if the position is not the same. For $e_y$, the $1$ is at position $y$, so if $(f({\bf x}))_y$ is also $1$, the error is $0$.

Interestingly, the formula also holds  when the assumption on $f(x)$ is relaxed to assuming that the elements are nonnegative and sum to $1$:
\begin{align}\Vert{e_y - f({\bf x})}\Vert_{1} &= \sum_j |(e_y)_j - (f(x))_j| \\
&= |(e_y)_y - (f(x))_y| + \sum_{j:j\neq y} |(e_y)_j - (f(x))_j| \\
&= |1 - (f(x))_y| + \sum_{j:j\neq y} |0 - (f(x))_j| \\
&= 1 - (f(x))_y + \sum_{j:j\neq y} (f(x))_j \\
&= 1 - 2(f(x))_y + \sum_j (f(x))_j \\
&= 2 - 2(f(x))_y
\end{align}


问:
Natural Deduction proof using basic rules only
I need some assistance solving what seems to be a very intuitive problem, but becomes tough when only using strict natural deduction and not assuming De Morgan laws.

Laws allowed: Implication, And, Or, MT, PBC, Copy Rule, Negation, Double Negation, Contradictions, law of excluded middle

I'm thinking it uses the law of excluded middle but I can't quite figure it out.

$$ \lnot(P \land \lnot Q), \; (\lnot P \to S) \land \lnot Q \;\;\; \text{premises} \tag{1} $$

$$ T \lor S \;\;\; \text{conclusion} \tag{2} $$

答: My proof here. Let me know if you see anything wrong with it, or if there's a more efficient way</p>


问:
Does the internal category forgetful functor have both adjoints?
Let $\mathcal{E}$ be a category with finite limits, and let $Cat(\mathcal{E})$ be the category of internal categories in $\mathcal{E}$, with forgetful functor $Ob : Cat(\mathcal{E}) \to \mathcal{E}$, sending each internal category in $\mathcal{E}$ to its $\mathcal{E}$-object of objects. A colleague of mine has conjectured/stated that this functor $Ob$ has left and right adjoints - is there a simple conceptual argument for this, and/or an appropriate reference in the literature somewhere?

答:
Not 100% sure that this works, but I would argue as follows.

Your friend's claim amounts to say that every object defines a discrete and a codiscrete internal category; discreteness is the easiest: the object of morphisms $C_1$ is just the object of objects, and all structure is trivial.

Codiscreteness can be argued similarly, abstracting the request that in the "maximaly connected groupoid" over a set there is exactly one arrow between any two objects.

It might help to recall that an internal category in $\cal E$ is just a truncated simplicial object $\Delta^o \to \cal E$ satisfying a Segal condition, so $Cat(\cal E)$ is a suitable category of functors; $Ob$ now is the "evaluation at $[0]$" functor.


答:
From our experience in the category of sets, we can see that the left adjoint should be a "discrete internal category" functor, and the right adjoint should be an "indiscrete internal category" functor. The discrete case is easier: we can define $L(X)$ to be the internal category $X\rightrightarrows X$ in which the source, target, identity-arrow, and composition morphisms are all the identity of $X$. One easily checks that this has the right universal property; this relies critically on the fact that the source and target morphisms have a common splitting, and wouldn't work at all for internal semicategories.

Now, what should the indiscrete internal category $R(X)$ be? If $X$ is a set, then the set of arrows in the indiscrete internal category is simply $X\times X$, so let's try defining $R(X)$ as $X\times X\rightrightarrows X$ in general. The source and target morphisms are simply the first two projections, while the identity-arrow morphism is the diagonal. The pullback $(X\times X)\times_X (X\times X)$ is just $X\times X\times X$, which the composition morphism the projection away from the middle factor.

Now given any morphism $f:A\to X$ in $\mathcal E$ and a category structure $(s,t):M\rightrightarrows A$ on $A$, we get a unique morphism $(f\circ s,f\circ t):M\to X\times X$ commuting with the source and target morphisms on either side, which one checks also respects the identity-arrow morphisms in both categories.

So both adjoints exist in the same manner as in the category of sets. It might be enlightening to observe that these adjoints are simply the adjoints to the evaluation-at-zero functor from the category of reflexive graphs internal to $\mathcal E$ to $\mathcal E$. Such evaluation functors always have adjoints on both sides, given by Kan extensions, if $\mathcal E$ admits enough limits and colimits, so one interesting thing that happens here is that the indexing diagram for reflexive graphs is so nice that no colimits are needed to construct the left adjoint to that evaluation functor, where normally some coproducts would be required. (This feels, morally, related to the fact that reflexive coequalizers are nicer than general coequalizers.)

The other interesting thing that happens is that the free and cofree reflexive graph on an object magically come equipped with canonical category structures. This is really a rather rare phenomenon, which is why you can't name any other algebraic structures for which the forgetful functor has a right adjoint.

Title: Does the internal category forgetful functor have both adjoints?

Body: Let $\mathcal{E}$ be a category with finite limits, and let $Cat(\mathcal{E})$ be the category of internal categories in $\mathcal{E}$, with forgetful functor $Ob : Cat(\mathcal{E}) \to \mathcal{E}$, sending each internal category in $\mathcal{E}$ to its $\mathcal{E}$-object of objects. A colleague of mine has conjectured/stated that this functor $Ob$ has left and right adjoints - is there a simple conceptual argument for this, and/or an appropriate reference in the literature somewhere?


问:
Uniform convergence of a functional series $\sum\limits _{n=1}^{\infty}\frac{x\sin(xn)}{n+x}$
Here's the question: Check series for uniform convergence: $$\sum\limits _{n=1}^{\infty}\frac{x\sin(nx)}{n+x}$$
a) $ x\in E_{1} = (0;\pi) ;$

b) $ x\in E_{2} = (\pi;+\infty) .$

For a) I used the fact that $ \left| \sum\limits_{n=1}^{N}x\sin(nx) \right| \le \left| x\sum\limits_{n=1}^{N}\sin(nx) \right| \le \left| \displaystyle{\frac{x}{\sin(\frac{x}{2})}} \right| \le \pi$ , that means we have bounded partial sums. Sequence $ \{\displaystyle{\frac{1}{n+x}}\} $ is monotone and uniformly goes to 0 for every value of $ x\in E_{1} $ as n goes to $\infty$. Thus, series is umiform convergent on $ E_{1} = (0;\pi) $ . Question is, can I perform the same thing setting partial sums as $ \left| \sum\limits_{n=1}^{N}\sin(nx) \right| $ and sequence as $ \{\displaystyle{\frac{x}{n+x}}\} $ which tells us that it converges to $1$ and finally saying that it converges not uniformly?

答:
For part (b), consider a sequence $x_n = \frac{\pi}{4n} + 2n\pi \in E_2$.  For $n < k \leqslant 2n$ we have

$$\frac{1}{\sqrt{2}} = \sin \left(\frac{\pi}{4}+ 2n^2 \pi \right) = \sin nx_n < \sin k x_n < \sin 2nx_n = \sin \left(\frac{\pi}{2}+ 2n^2 \pi \right) = 1$$

Thus,

$$ \left|\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} \right| =\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} > n \cdot \frac{x_n}{2n + x_n}\cdot \frac{1}{\sqrt{2}} = \frac{2n^2\pi + \frac{\pi}{4}}{\sqrt{2}(2n + 2n\pi + \frac{\pi}{4n})} $$

Since, the RHS tends to $+\infty$ as $n \to \infty$, the series cannot be uniformly convergent on $E_2$, by violation of the uniform Cauchy criterion.

Edit (9/16/2022):

The argument above is flawed because with $x_n = \frac{\pi}{4n} + 2n\pi$, we should have $\frac{\pi}{4} +2n^2\pi < kx_n \leqslant \frac{\pi}{2} + 2n^2\pi + 2n^2\pi$ and we don't have $\sin kx_n > 0$ for all $n< k\leqslant 2n$.

It should be amended as follows.

Consider a sequence $x_n = \frac{\pi}{4n} + 2\pi \in E_2$.  For $n < k \leqslant 2n$ we have $\frac{\pi}{4} < \frac{\pi k}{4n}\leqslant \frac{\pi}{2}$ and

$$\frac{1}{\sqrt{2}} = \sin \left(\frac{\pi}{4}\right) < \sin \frac{\pi k}{4n} =\sin \left(\frac{\pi k}{4n}+ 2k\pi\right) = \sin kx_n  < \sin \left(\frac{\pi}{2} \right) = 1$$

Thus,

$$ \left|\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} \right| =\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} > n \cdot \frac{x_n}{2n + x_n}\cdot \frac{1}{\sqrt{2}} = \frac{2n\pi + \frac{\pi}{4}}{\sqrt{2}(2n + 2\pi + \frac{\pi}{4n})} $$

Since the RHS tends to $\frac{\pi}{\sqrt{2}}\neq 0$ as $n \to \infty$, the series cannot be uniformly convergent ...


问:
Does there exist an English translation of Distler's paper on solving polynomials by radicals?
The Radiroot package for GAP by Andreas Distler solves by radicals polynomials with solvable Galois group. The package uses the algorithm described in Distler's paper "Ein Algorithmus zum Lösen einer Polynomgleichung durch Radikal". As you would guess from the title, the paper is in German, which I cannot read. Is there an English (or Russian) translation of this paper?</p>

答:
I've checked around, but can't seem to find a translated version of the paper, be it English or Russian.  I did Google it. As a last resort, you can try Google's "translate" option: it rendered a decent translation, though formatting suffers.

Perhaps your best bet would be to contact Andrew Distler directly?


问:
Prove that $\lim\limits_{x\to+\infty}f'(x)=\lim\limits_{x\to+\infty}f''(x)=0$

Given $f \in C^3(\mathbb{R}).$ Suppose that
$\lim\limits_{x\to+\infty}f(x)=a \in \mathbb{R}$ and
$\lim\limits_{x\to+\infty}f'''(x)=0$. Prove that
$\lim\limits_{x\to+\infty}f'(x)=\lim\limits_{x\to+\infty}f''(x)=0$.

My solution: Since $\lim\limits_{x\to+\infty}f(x)=a$, there exists $M>0$ that for all $x, x' \in \mathbb{R}$, if $x,x'>M$ then $f(x), f(x') \in \left(a-\varepsilon^2/2, a+\varepsilon^2/2\right)$. This yields
$$\vert f(x)-f(x') \vert < \varepsilon^2.$$

Therefore $\frac{\vert f(x)-f(x') \vert}{\varepsilon} < \varepsilon$. So, for all $x>M$, $x'=x+\varepsilon$ we have
$$0\le\frac{\vert f(x+\varepsilon)-f(x) \vert}{\varepsilon} < \varepsilon.$$
Let $\varepsilon \to 0$ we have
$$0 \le f'(x) \le 0 \text{ or } f'(x)=0.$$

That means $\lim\limits_{x\to+\infty}f'(x)=0$.

Doing similarly as above we have
$$\lim\limits_{x\to+\infty}f''(x)=0.$$

My question: $\lim\limits_{x\to+\infty}f'''(x)=0$ is really necessary or not (because I don't use it in my proof)?

答:
First, consider
$$
f(x)=\frac{\sin(x^2)}{x}.
$$
As $x\rightarrow\infty$, this approaches $0$.  On the other hand,
$$
f'(x)=\frac{2x^2\cos(x^2)-\sin(x^2)}{x^2}=2\cos(x^2)-\frac{\sin(x^2)}{x^2}
$$
does not have a limit as $x\rightarrow\infty$.  Therefore, the condition on $f'''$ is necessary for the result.

The problem with your setup is that you never specify when you're quantifying $\varepsilon$.  You're actually quantifying it before computing $M$, so we'll write $M=M(\varepsilon)$.  Next, you consider
$$
\frac{|f(x+\varepsilon)-f(x)|}{(x+\varepsilon)-x}.
$$
At this point, you let $\varepsilon$ go to $0$, but here's your problem, as $\varepsilon$ goes to $0$, $M(\varepsilon)$ may go to infinity, so you don't get a bound on the derivative when $\varepsilon$ is too small, so the squeeze theorem cannot be applied.


问:
How can I simplify this lerp arithmetic to avoid floating point precision errors?
I am making a "cinematic camera" that pans around a scene. Each step has a position for the camera and a point that the camera should be looking at. For instance:

```<code>1. position = (3, 2, 1), facing = (0, 0, 0)
2. position = (6, 5, 4), facing = (9, 9, 9)
</code>```

The camera should start at (3, 2, 1) and be looking at the point (0, 0, 0). By the end of the panning the camera should be positioned at (6, 5, 4) and be looking at the point (9, 9, 9).

I've written some C# to do this:

```<code>public static IEnumerable<(Vector3, Vector3)> CinematicCamera(params (Vector3 location, Vector3 facing, double time)[] points) {
    if (points.Length < 2) {
        throw new ArgumentException("Need at least 2 points.");
    }

    List<(Vector3, Vector3)> returnMe = new List<(Vector3, Vector3)>();

    for (int i = 1; i < points.Length; i++) {
        int numSteps = (int)(points[i].time * 20); // camera runs at 20 frames per second
        double startDistance = Distance(points[i - 1].location, points[i - 1].facing); // distance the camera starts from the point it's looking at
        double endDistance = Distance(points[i].location, points[i].facing); // distance the camera ends from the point it's looking at
        Vector3 startDirection = points[i - 1].location - points[i - 1].facing; // get the start directional vector from the location point and facing point
        Vector3 endDirection = points[i].location - points[i].facing; // get the end directional vector from the location point and facing point

        for (int stepNum = 0; stepNum < numSteps; stepNum++) {
            double progress = (double)stepNum / numSteps; // 0 to 1 based on current step, used for lerp
            double distance = Lerp(startDistance, endDistance, progress); // lerp the distance from the previous point to the next point
            Vector3 currentDirection = Lerp(startDirection, endDirection, progress); // lerp the directional vector

            Vector3 lookAt = Lerp(points[i - 1].facing, points[i].facing, progress); // new point to look at
            Vector3 position = lookAt + currentDirection * distance / Length(currentDirection); // new position

            returnMe.Add((position, lookAt));
        }
    }

    return returnMe;
}

// helper methods
private static double Distance(Vector3 a, Vector3 b) => Math.Sqrt(Math.Pow(a.X - b.X, 2) + Math.Pow(a.Y - b.Y, 2) + Math.Pow(a.Z - b.Z, 2));
private static double Length(Vector3 v) => Math.Sqrt(v.X * v.X + v.Y * v.Y + v.Z * v.Z);
private static double Lerp(double start, double end, double amount) => start + (end - start) * amount;
private static Vector3 Lerp(Vector3 start, Vector3 end, double amount) => new Vector3(Lerp(start.X, end.X, amount), Lerp(start.Y, end.Y, amount), Lerp(start.Z, end.Z, amount));
</code>```

Vector3 is a custom data type that uses double to store x, y, and z. This all works, but the camera is a bit jittery due to floating point imprecision. I graphed some of the points in Excel and you can see the jitter in the graph:

I'm considering switching the entire implementation from double to decimal (double in C# uses 8 bytes while decimal uses 16), but I'm not sure that really addresses the root of the problem. How can I simplify this so that I run into less floating point weirdness?

答:
I accidentally fixed this. The problem was not floating point inaccuracy, it was actually this line here:

```<code>double progress = (double)stepNum / numSteps; // 0 to 1 based on current step, used for lerp
</code>```

It should be

```<code>double progress = (double)stepNum / (numSteps - 1); // 0 to 1 based on current step, used for lerp
</code>```

Body:
I accidentally fixed this. The problem was not floating point inaccuracy, it was actually this line here:

double progress = (double)stepNum / numSteps; // 0 to 1 based on current step, used for lerp

It should be

double progress = (double)stepNum / (numSteps - 1); // 0 to 1 based on current step, used for lerp


问:
Understanding Fubini's theorem in stochastic calculus
I am confronted with the following dilemma: what are the conditions under which Fubini's theorem is applicable in order to exchange Integrals and Expectations. Let me have a stochastic process $X = \{X_t\}_{t \geq0}$. When I am allowed to write the following equality?
$$\mathbb{E} \left[ \int_0^t f(X_s) ds \right] = \int_0^t\mathbb{E} \left[  f(X_s) \right]ds$$
Is it sufficient that $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous?

答:
Continuity is clearly not enough. Take:

$f (x) = x$;

the underlying measure space for the process as $[0,1]$ with the Lebesgue measure;

$X_s (\omega) = g(s, \omega)$ for some measurable function $g$.

Then you are asking whether

$$\int_0^1 \int_0^t g(s, \omega) \ \text{d}s \ \text{d} \omega = \int_0^t  \int_0^1 g(s, \omega) \ \text{d}\omega \ \text{d} s,$$

which is the usual Fubini theorem for $g$, and thus still needs additional assumptions (positivity or integrability).

The answer is going to be boring; the equality holds if $f$ is nonnegative or integrable (in particular bounded). The later condition reads:

$$\int_0^t \mathbb{E} (|f(X_s)|) \ \text{d} s < +\infty.$$


答:
Thanks to @LinAlg, I got the answer. I'm writing it for 2 classes and can be generalized further.

Consider the predictions $ f({\bf x}) = [\epsilon, 1-\epsilon]$ and ground truths $e_y = [1, 0]$. The mean absolute error will be

$$MAE = |1-\epsilon| + |-1+\epsilon| = 2 - 2\epsilon$$

This will be same even if $e_y = [0, 1]$.

Title: Can't understand Mean Absolute Error (MAE) formulation

Body:
I'm reading Taylor Cross Entropy Loss paper and came across the formulation of Mean Absolute Error (MAE), which is described as following:

$$
\mathcal{L}_{MAE}(f({\bf x}), y) = \Vert{e_y - f({\bf x})}\Vert_{1} = 2 - 2f_y({\bf x})
$$

As mentioned in the linked paper's section 3.1, $e_y$ is one hot encoded vector having same dimension as $f({\bf x})$.

What I don't understand is that from where does the constant 2 come from in above formulation? I'm referring to this formulation if MAE on Wikipedia.

Any hint/reference would be appreciated. Thanks.

Edit

As requested in the comment, here is the context of the above formulation.

The linked paper talks about training a deep neural network for k class classification. ${\bf x}$ is a feature vector (e.g. image of cat) and $y$ is ground truth label.

The neural network is represented as an unknown complex function $f$.

$e_y$ is one hot encoded vectors of ground truths. For example, if k=2 (cat and dog), then $e_y = [0, 1]$ for cat and $e_y = [1, 0]$ for dog.

$f({\bf x})$ is what predicted by neural network. It can be probabilities for each class. For example, a well trained neural network will output $f({\bf x}) = [0.05, 0.95]$ for an image of cat.

The mean absolute difference between ground truth labels and the predicted outputs for all images is what I refer to MAE.


问:
Velleman's Proof Designer problem 44
This is an exercise I didn't solve from the group of exercises present on Velleman's site that should be solved using Proof Designer (the little program that comes with the book).

Theorem:

If $\forall \mathcal{F} (\cup \mathcal{F} = A \rightarrow A \in \mathcal{F})$, then $\exists x (A=\{x\})$.

Before addressing my line of reasoning and the tentative proof, just few words on this specific problem. I find it interesting because it shows me that I still have problems figuring (and writing) properly existence results. About this, I would say that no matter what book on proof writing you take, this point is usually quickly considered and the examples are trivial, so not particularly useful for real mathematics. Probably this is related to the fact that existence theorems are in general the "hard ones" to be proven and to be thaught (I guess that they are the most difficult to be grasp by students), but I really don't know and I would like to know the opinions of experts.

Line of Reasoning:

I think the idea is that $x=\emptyset$ and $A=\{\emptyset\})$. In this way, what we have to assume should stand. Indeed, the antecedent should "bind" the nature of the family of sets and the consequence would follow.

Tentative Proof:

Let x be the empty set $\emptyset$. Thus, $\forall \mathcal{F} (\cup \mathcal{F} = \{\emptyset\} \rightarrow \{\emptyset\} \in \mathcal{F})$. Take $\mathcal{F}=\{\emptyset,\{\emptyset\}\}$. Then, we have $\cup \{\emptyset,\{\emptyset\}\} = \{\emptyset\} \rightarrow \{\emptyset\} \in \{\emptyset,\{\emptyset\}\}$, and the result is established.

PS: As I pointed out in another post, in general I try to be more consistent grammatically speaking when I write down proofs, and I avoid the use of logical symbols. I tend to do it when I have doubts about the content of the problem and here... I have no clue at all, so forgive it. Any feedback is more than welcome.

答:
Your tentative proof is the wrong direction: If $A$ is a singleton (in fact a very specific singleton) then the antecedent holds.

Under the assumption of  $\forall \mathcal{F} (\cup \mathcal{F} = A \rightarrow A \in \mathcal{F}) $, first show $\exists x(x\in A)$. In other words: Show that $A=\emptyset$ implies $\neg\forall \mathcal{F} (\cup \mathcal{F} = A \rightarrow A \in \mathcal{F}) $ or equivalently $\exists \mathcal{F} (\cup \mathcal{F} = A \land A \notin \mathcal{F}) $. You can do this by specifically exhibiting such $\mathcal F$.

Next show $(x\in A\land y\in A)\rightarrow x=y$. Again, exhibit $\mathcal F$ with $\cup \mathcal{F} = A \land A \notin \mathcal{F}$.

You should already know that $x\in A$ and $(x\in A\land y\in A)\rightarrow x=y$ implies $A=\{x\}$.

Title: Velleman's Proof Designer problem 44


问:
If $\psi=x\sin(\frac{1}{x})$, and $f$ is integrable, is then $f\circ \psi$ also integrable?
Let
$$
\psi(x)=
\left\{
\begin{array}{cll}
x \sin\Big(\dfrac{1}{x}\Big) & \text{if} & x\in (0,1],\, \\
0 & \text{if} & x=0,
\end{array}
\right.
$$
and let $f:[-1,1]\rightarrow \mathbb{R}$ be Riemann integrable.

How can I show that $f\circ \psi$ is Riemann integrable?

I have several theorems in my book that could work if $\psi$ were a $C^1$ diffeomorphism or a homeomorphism with inverse satisfying Lipschitz condition. But $\psi$ is clearly neither of those. Any help with a zero set argument or reverting to definitions?

答:
i) If $\psi:[\alpha,\beta]\to [a,b]$ is $C^1$ and $\psi'$ has only finitely many zeros, then $f\circ \psi$ is RI on $[\alpha,\beta]$ for every $f$ RI on $[a,b].$ I'll take this one as known.

ii) Let's generalize: Now suppose $\psi:[\alpha,\beta]\to [a,b]$ is $C^1$ and $\psi'$ has finitely many zeros in every closed interval contained in $(\alpha,\beta).$ Then $f\circ \psi$ is RI on $[\alpha,\beta]$ for every $f$ RI on $[a,b].$ To prove this, let $f$ be RI 0n $[a,b].$ Let $\epsilon>0$ be small. Then from i) we have $f\circ \psi$ RI on $[\alpha+\epsilon,\beta-\epsilon].$ Letting $\epsilon\to 0^+,$ we get $f\circ \psi$ is RI on $[\alpha,\beta]$ for every $f$ RI on $[a,b].$

Corollary: Suppose $\psi:[\alpha,\beta]\to [a,b]$ is real analytic on $(a,b).$ Then $f\circ \psi$ is RI on $[a,b]$ for all $f$ RI on $[a,b].$ Proof: If $\psi$ is constant there is nothing to prove. If $\psi$ is nonconstant, then $f'$ is real analytic and not $\equiv 0$ on $(a,b).$ By the identity principle, $\psi$ has finitely many zeros in every closed interval contained in $(\alpha,\beta).$ From ii) we then have the desired result.

The corollary gives a quick non computational proof of the problem at hand, since $x\sin(1/x)$ is real analytic on $(0,1).$


问:
Formula for radius of circles on vertices of regular polygon
TLDR: Given a square, a number of circles, a rotation angle, and some spacing factor, how can I determine the largest possible radius of circles arranged on the vertices in a regular polygon such that the circles do not exceed the bounds of the square?

I am trying to determine a formula or algorithm for calculating the radius of circles situated on the vertices of regular polygons with 1 to n sides.

I have tried the following with inconsistent results.

```<code>2 * a * tan(π / n) / x
</code>```

Where x is a sizing factor and a is the apothem. For example say I wanted the spacing between circles to be such that you could fit another circle on the midpoint of the side, x could be 4. Or if I wanted the circles to be touching x could be 1. But for n = 3 the results look quite different than say for 6 in that the spacing between the circles is too small/big respectively.

As for the apothem, it's not clear how I would calculate that.

For n = 1 or 2 I would have to handle it differently. For 1 I'd just want a circle in the centre of the available space, and for 2 they would be arranged in a line. Is there a more general formula I could use?

Something like the circles on the vertices of the hexagon in the image below, where I could adjust the radius of the circles as needed.

Further complication is that I calculate the coordinates of the vertices based on a given square and an angle to offset the resulting vertices, but I reduce the size of the resulting polygon based on the radius of the circles to make sure the resulting circles do not exceed the bounds of the square.

Here's a few imperfect drawings I made for 1–3:

The circles don't have to fill the square necessarily; I want to be able to adjust the spacing between them by reducing the radius.

I'm a bit stuck on a circular dependency (no pun intended) since I think I need the positions of the vertices to calculate the circle radius, but I need the circle radius to determine the positions of the vertices so that the circles do not exceed the size of the square.

答:
I'm not sure how mathematical this is but instead of placing the circles so their centres are on the vertices of the polygon (which creates a cyclical dependency between the radius of the circles and the polygon vertices) I placed the circles so that the polygon vertex lies on the circumference of the circle and I used linear interpolation relative to the centre of the polygon to calculate the coordinates of the centres of the circles. For the case of n == 1 I still need to handle it a bit differently but other than that it works nicely.

I also used the distance from the vertices to the centre to adjust the radius of the circles to adjust the spacing.


问:
Solve the differential equation: $xy'' + y' = 0$
Title of the question states it. I need to solve $xy'' + y' = 0$. Any pointers? I know it's easy but I struggle with differential equations.... Thanks in advance.</p>

答: $$xy''+y'=(xy')'=0\to xy'=c\to y'=\dfrac cx\to y=c\log x+d.$$


问:
Elliptic curve prime power isogenies
Let $E/K$ be an elliptic curve, where $K$ is a number field.
Let $p>2$ be a prime number.

Is there any bound on $n$ such that $E$ has a $K$-rational cyclic $p^n$-isogeny?

Can it have such isogeny for infinitely many $n$?

答:
If $\phi\colon E\to E'$ is a rational cyclic $p^n$-isogeny with kernel $E[\phi]$, then $E[\phi]$ is a $G_K = \mathrm{Gal}(\overline K/K)$ stable subgroup of $E[p^n]$. It follows that $E[p^n]$ is reducible as $G_K$-module.

Assume that $E$  is not CM. In particular, by a theorem of Serre/Shafarevich, the Tate module $\varprojlim_n E[p^n]$ is irreducible as a $G_K$-module. It follows that $E[p^n]$ must be irreducible when $n$ is large enough. In particular, $E$ cannot have a rational $p^n$-isogeny for infinitely many $n$.

As for a uniform bound depending on $K$, one probably exists, although it might be hard to prove - hopefully someone more knowledgeable than me can answer!

Except when $K = \mathbb Q$, the analogous question asking for a uniform bound depending only on  for the largest prime  for which there exists a -isogeny is an open problem!

Title: Elliptic curve prime power isogenies


问:
Limit of curvilinear integral
Let $I(R)=\int_{\gamma_R}(1+x_1^2+x_2^2)^{-1/2}\cos(x_2)ds$, where $\gamma_R=\{(x_1,x_2):x_1^2+x_2^2=R^2,x_1>0\}$. Compute $\lim_{R \to +\infty}$$I(R)$. I tried to solve it in this way: I use parameterization $x_1=R\cos\phi, x_2=R\sin\phi, \phi \in (-\pi/2,\pi/2)$ and I get: \begin{equation} I(R)=\frac{R}{(1+R^2)^{1/2}}\int_{-\pi/2}^{\pi/2}\cos(R\sin\phi)d\phi
=\frac{2R}{(1+R^2)^{1/2}}\int_{0}^{\pi/2}\cos(R\sin\phi)d\phi   \end{equation}
Now for $R \to +\infty$ :
\begin{equation}\frac{2R}{(1+R^2)^{1/2}}\sim2 \end{equation}
How could I study the integral to conclude the exercise?

答: Hint: In your last integral, let $\phi = \arcsin t.$ You will then have an integral of the form $\int_0^1 f(t)\cos(Rt)\,dt.$ Think about the Riemann Lebesgue lemma.

Title: Limit of curvilinear integral


问:
Equivalent condition for being Positive element of a C*-algebra
This is a problem from the book "A course in functional analysis" by John B. Conway (Chapter 8 Exercise 8.9).

If $\mathcal A$ is an unital $\textit C^*$-algebra. For $a\in \mathcal A$, $a\ge 0$ if and only if $f(a)\ge 0$ for every state $f$.

One side is trivial from definition. The problem is to prove the other direction. If I can prove $a$ is self-adjoint then I am done, because for every $\lambda$ in spectrum of $a$ there is a state such that $f(a)=\lambda$.
I couldn't show it is self adjoint.

答:
This is basically Theorem 3.4.3. in Murphy's book.

Let $\pi : \mathcal{A} \to \mathbb{B}(H)$ be a faithful nondegenerate representation of $\mathcal{A}$ on a Hilbert space $H$. For a fixed unit vector $x \in H$ consider the functional $$f : A \to \Bbb{C}, \quad f(b)=\langle \pi(b)x,x\rangle.$$
It is easy to see that $f$ is a state so we have $0\le f(a) = \langle \pi(a)x,x\rangle$. In particular, we see that $$\langle \pi(a)x,x\rangle
\ge 0, \quad \forall x \in H$$
which implies that $\pi(a)$ is self-adjoint and then $\pi(a) \ge 0$.

Now we have $\pi(a) = \pi(a)^* = \pi(a^*)$ so by faithfulness we get $a = a^*$. You can also get $a \ge 0$ automatically if you recall that $\pi(\mathcal{A}_+) = \pi(\mathcal{A})_+$ holds.


问:
Generating all k-regular graphs by switching edges
I want to randomly place $k$ zeroes in each row of a $n \times n$ matrix, such that the resulting matrix is symmetric and no zeroes are on the main diagonal of the matrix.

This problem is equivalent to generating random $k$-regular graphs on n vertices. I have already proven, that this is not possible for odd k and odd n. And I have already found a way to generate a k-regular graph for the other case. I wondered if it is possible to reach every other $k$-regular graph by switching edges. Therefore I want to know if the following can be proven or if there is a counter example.

Let $G$ and $G'$ be $k$-regular graphs on $n$ vertices. Is it possible to change one graph to the other by just switching edges while still having a regular graph after each switch?
(The edges $AB$ and $CD$ get changed to $AC$ and $BD$)

I have already proven that you can switch edges of $G'$ such that a single vertex has the same adjacent vertices as in $G$, but then I haven't found a way to switch the edges in a way that preserves the adjacent vertices of the first m vertices and results in m+1 vertices which having the same adjacent vertices in $G'$ and $G$.

答:
More is true: as long as there is at least one graph with a given degree sequence, all graphs with a given degree sequence are connected by the edge swap operation. You are only interested in the regular case, but that isn't any easier than the general case to prove.

This result is due independently to Havel and Hakimi, but most people are only aware of the Havel–Hakimi algorithm for determining if a sequence is a degree sequence of some graph. That algorithm is the direct corollary of this result.

Suppose two graphs $G$ and $H$ have the same degree sequence: we can label their vertices $v_1, v_2, \dots, v_n$ such that $\deg_G(v_i) = \deg_H(v_i)$ for $i=1,2,\dots,n$. Instead of explaining how to turn $G$ into $H$, I will explain how to turn $G$ and $H$ both into the same graph. Since the operations are reversible, we can get from $G$ to that graph to $H$. Here's how...

Without loss of generality, let's suppose that the vertices are sorted in decreasing order of degree (in both graphs). Then we can:

Perform swaps in both graphs so that $v_1$ is adjacent to $v_2, v_3, \dots, v_{d+1}$ in both graphs, where $d = \deg_G(v_1) = \deg_H(v_1)$.

Use this algorithm to perform swaps on $G - v_1$ and $H - v_1$ to turn them both into the same graph. This will have the effect of turning $G$ and $H$ into the same graph. We might need to reorder the vertices when we do this, so that the degrees of $v_2, \dots, v_n$ in $G-v_1$ and $H-v_1$ are again in descending order.

Of course, we need to prove that step 1 is always possible.

Suppose that in one of the graphs ($G$, without loss of generality) $v_1$ is not yet adjacent to some vertex $v_i \in \{v_2, \dots, v_{d+1}\}$. Then it must be adjacent to some vertex $v_j$ not in this set. Vertices are sorted by degree, so $\deg(v_i) \ge \deg(v_j)$.

Vertex $v_i$ has $\deg(v_i)$ neighbors among $\{v_2, v_3, \dots, v_n\}$. Vertex $v_j$ only has $\deg(v_j)-1 < \deg(v_i)$ neighbors in that set, since it is also adjacent to $v_1$. (If $v_i v_j$ is an edge, this contributes $1$ to both degrees, so we can ignore it.) Therefore there is a fourth vertex $v_k$ adjacent to $v_i$ but not $v_j$.

Now perform a swap: replace edges $v_1 v_j$ and $v_i v_k$ by $v_1 v_i$ and $v_j v_k$. This increases the number of neighbors of $v_1$ in the set $\{v_2, \dots, v_{d+1}\}$. If we repeat this operation, we can put $G$ and $H$ both in the form we wanted.


问:
"Spread-Out" Partitions
Firstly, the given definition of spread-out is

"A partition is said to be 'spread-out' if all the parts are distinct and no two parts are consecutive numbers."

I am trying to work out how many spread-out partitions of 21 are there into 4 parts? I believe the answer is 7, but I only achieved this by brute force, and hence wondering if there is a easier way and more general formula of working this out.

答:
Let $f(n,k,m)$ be the number of spread-out partitions of $n$ into $k$ distinct parts of size at most $m$.  By conditioning on whether part $m$ appears, we find that
$$f(n,k,m) = f(n-m,k-1,\min(m-2,n-m)) + f(n,k,m-1)$$
with obvious boundary conditions.
We want to compute $f(21,4,21)$, which turns out to be $6$.  The corresponding partitions are
$$
12+5+3+1\\
11+6+3+1\\
10+7+3+1\\
10+6+4+1\\
9+7+4+1\\
9+6+4+2\\
$$

Title: "Spread-Out" Partitions


答:
You are still quite far from proving the theorem. The issue seems to be that you misunderstood what the theorem is saying. I will try to explain it with a minimum of logical symbols, as you said you prefer that.

We are given a set $A$. First, let us investigate the premise:

$$\forall \mathcal F: \cup \mathcal F = A \to A \in \mathcal F$$

One could say that this expresses that $A$ is "union-irreducible": when it is the union of any collection of sets $\mathcal F$ (i.e. $\cup\mathcal F = A$), then it must be the case that one of these sets in $\mathcal F$ already is $A$.

It is important that we have the universal quantifier $\forall$ here, because we want to say that $A$ is "irreducible" under all unions, not just some particular ones.

Now let us turn to the conclusion:

$$\exists x: A = \{x\}$$

It does not say anything but: $A$ has precisely one element (formally: "there is an $x$ such that $x$ is the only element of $A$).

Thus, the statement of the theorem becomes: Any union-irreducible set $A$ consists of precisely one element.

In particular, we are not allowed to pick a set $A$, or, equivalently, the element $x \in A$. The theorem has to be proved for all union-irreducible sets.

For completeness, I will indicate a proof of the theorem. Let $A$ be a union-irreducible set.

We have the set $\mathcal F = \{\{x\}:x \in A\}$, comprising a singleton $\{x\}$ for each $x \in A$. It is trivial to see that $\cup\mathcal F = A$.

So now we need the conclusion to hold, i.e. $A \in \mathcal F$. But this means that there must exist an $x\in A$ such that $A = \{x\}$. The theorem follows.

Title: Velleman's Proof Designer problem 44

Body:
This is an exercise I didn't solve from the group of exercises present on Velleman's site that should be solved using Proof Designer (the little program that comes with the book).

Theorem:

If $\forall \mathcal{F} (\cup \mathcal{F} = A \rightarrow A \in \mathcal{F})$, then $\exists x (A=\{x\})$.

Before addressing my line of reasoning and the tentative proof, just few words on this specific problem. I find it interesting because it shows me that I still have problems figuring (and writing) properly existence results. About this, I would say that no matter what book on proof writing you take, this point is usually quickly considered and the examples are trivial, so not particularly useful for real mathematics. Probably this is related to the fact that existence theorems are in general the "hard ones" to be proven and to be thaught (I guess that they are the most difficult to be grasp by students), but I really don't know and I would like to know the opinions of experts.

Line of Reasoning:

I think the idea is that $x=\emptyset$ and $A=\{\emptyset\})$. In this way, what we have to assume should stand. Indeed, the antecedent should "bind" the nature of the family of sets and the consequence would follow.

Tentative Proof:

Let x be the empty set $\emptyset$. Thus, $\forall \mathcal{F} (\cup \mathcal{F} = \{\emptyset\} \rightarrow \{\emptyset\} \in \mathcal{F})$. Take $\mathcal{F}=\{\emptyset,\{\emptyset\}\}$. Then, we have $\cup \{\emptyset,\{\emptyset\}\} = \{\emptyset\} \rightarrow \{\emptyset\} \in \{\emptyset,\{\emptyset\}\}$, and the result is established.

PS: As I pointed out in another post, in general I try to be more consistent grammatically speaking when I write down proofs, and I avoid the use of logical symbols. I tend to do it when I have doubts about the content of the problem and here... I have no clue at all, so forgive it. Any feedback is more than welcome.


问:
Are $\mathbb Z\times \mathbb Z/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$ as groups?
Is $(\mathbb Z\times \mathbb Z)/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$,

where $4\mathbb Z\times 5 \mathbb Z$ is given as a subgroup of $\mathbb Z\times \mathbb Z$?

My feelings says that I need to find well defined $f:\mathbb Z\times \mathbb Z\to \mathbb Z$ which has a kernel $4\mathbb Z\times 5\mathbb Z$

So using first isomorphism theorem I can show the isomorphism.

I could not find the map and how can I show it in different way?

答:
There are a number of answers explaining why your isomorphism isn't true, but I am going to try to answer to the question maybe you wanted yo ask: let $A = \mathbb{Z}\times \mathbb{Z}$, and $B = \{(4x,5x)\in A\mid x\in \mathbb{Z}\}$. Note that $B$ is not the subgroup $4\mathbb{Z}\times 5\mathbb{Z}$, and is not even isomorphic to it since it is instead isomorphic to $\mathbb{Z}$.

Then indeed $A/B\simeq \mathbb{Z}$, because $4$ and $5$ are coprime. Explicitly, the surjective morphism $A\to \mathbb{Z}$ given by $(a,b)\mapsto 5a-4b$ has kernel $B$.


问:
Probability of rolling at least one of two specified numbers using 6-sided dice
What is the formula - in simple language - to calculate the probability of rolling at least one of the two specified numbers using three 6-sided dice?

Let's say, the next roll of the three 6-sided dice needs to include at least one 5 or 6 (or both). In other words, I win if any of these show up at least once on any of the three dice on the next roll.

I do know the result should be 0.7037 but I could not reverse engineer it. There are 216 possible outcomes. I have a 0.33 probability of rolling either a 5 or a 6 with a single die, but I have three dice.

答:
Our desired event is complementary to the event that none of the two specified number appears. Hence required probability is
$$1-\frac{4^3}{6^3}=\frac{19}{27}\approx 0.7037$$


答:
For questions like this (about "at least one...") a good trick is to compute your probability of failure.

For you to fail each die must come up 1,2,3, or 4. Since there's a $2/3$ probability of that happening for each individual die, the probability that it happens for all three of them is $(2/3)^3 = 8/27$. Thus, finally, the probability of you succeeding is $1-(2/3)^3 = 19/27 \approx 0.7037\dots$.

Title: Probability of rolling at least one of two specified numbers using 6-sided dice

Body:
What is the formula - in simple language - to calculate the probability of rolling at least one of the two specified numbers using three 6-sided dice?

Let's say, the next roll of the three 6-sided dice needs to include at least one 5 or 6 (or both). In other words, I win if any of these show up at least once on any of the three dice on the next roll.

I do know the result should be 0.7037 but I could not reverse engineer it. There are 216 possible outcomes. I have a 0.33 probability of rolling either a 5 or a 6 with a single die, but I have three dice.


问:
In how many cases can a matrix be singular, for rotation, projection or reflection matrix?
I'm currently doing an old exam (linear algebra), my real one is in three days. There was a question in it and I wanted to make sure that my way of thinking is correct. The question is:

When one expresses the following maps as matrix; in how many cases are the matrices non-singular?

a rotation in $\mathbb{R}^2$,

an orthogonal projection in $\mathbb{R}^2$ onto a line,

an oblique projection in $\mathbb{R}^2$ onto a line,

the reflection in $\mathbb{R}^2$ in the origin.

My answer:

A rotation matrix is non-singular since it is invertible.

Now I know that all projection matrices except the identity matrix are singular. The identity matrix is an orthogonal projection since: $I^2 = I$ and $I^T = I$. This implies that an orthogonal projection can be non-singular in the case that the projection matrix is the identity matrix.

A projection matrix $P$ is oblique $\iff$ $P^2 = P$ and $P \ne P^T$. I figured that, since the only matrix that is a projection and is non-singular is $I$, and since $I = I^T \implies \not\exists P : P^2 = P$ and $P^T \ne P$.

A reflection is invertible and hence non-singular.

Which concludes that in three cases the matrix can be non-singular.

I'm very curious whether or not I'm missing something and I would appreciate if someone could point out is that is the case. When doing the homework problems I have never encountered a question like this before.
Thank you.

Edit: It is a multiple choice question where I can choose between 1,2,3,4 and 5 possibilities where a matrix can be non-singular. So I don't have to prove it, however, im curious if my way of thinking is correct. (Why there is a fifth option in the answer I don't know)

答:
Answering my own question.

At option number 2 in the question, that ask if the matrix for an orthogonal projection in $\mathbb{R}^2$ onto a line is non-singular. Indeed, the identity matrix is a projection, however not onto a line. This rules out the case of the identity matrix and hence no such matrix can be invertible. Concluding that there are two cases in which the matrix is non-singular, namely, the rotation and the reflection.


问:
Relevance of a statement about monster models
I was reading this question and I have trouble understanding what the second statement really says. I know almost nothing about monster models and have never worked with them, so I should probably read some more introductory material but I still would like to understand that bit.

I copy below this second statement I'm speaking about. $p(x)$ is a not necessarily complete type and $A$ and $B$ are from what I understand submodels of this monster model (is $B$ supposed to be an elementary submodel?).

Let $A\subset B$ and $p(x)\subset L(A)$ suppose that for any solution $a$ of $p(x)$ the complete type over $B$, $tp(a/B)$ is isolated. Show that $p(x)$ is isolated.

Let me try to express this in a language which is more familiar to me. Let $M$ be the monster model, let $M_A$ be the same structure but where we add one constant per element of $A$ to the language and let $\DeclareMathOperator{\Typ}{Typ}\Typ_{M_A}(n)$ be the space of $n$-types of the theory of $M_A$. Then from what I understand, $p(x)$ is a closed set in $\Typ_{M_A}(1)$ and $tp(a/B)$ is a point of $\Typ_{M_B}(1)$. We have a function $\Typ_{M_B}(1) → \Typ_{M_A}(1)$ and the statement says that if all the points in the inverse image of $p(x) ⊆ \Typ_{M_A}(1)$ are isolated (open), then $p(x)$ is open itself. This follows from the fact that $\Typ_{M_B}(1) → \Typ_{M_A}(1)$ is surjective and continuous: it is then a quotient.

Is what I said above correct? If so, I don't understand what is the significance of this property. From what I understand, when working with a monster model, we place everything inside of it and it should stay "implicit" but ease the arguments. I have never worked with a monster model, so I don't really know what is going on here: what does the above property says about an embedding $A⊆B$ between two structures when we forget that there is a monster model?

答:
There are a couple of issues raised in your question that I want to address, so this answer might seem a bit unfocused.

1. Some comments on terminology and notation:

I prefer not to use the word "isolated" to describe a partial (possibly incomplete) type. I would rather call such a type "equivalent to a formula" or "finitely axiomatizable" or even "clopen". I prefer to reserve the word "isolated" for complete types (since when a complete type is equivalent to a formula, it is an isolated point of the Stone space). I used the word "isolated" in my answer to the question you linked to because I wanted to follow the OP's terminology - I probably should have also made this comment there.

The standard notation for the Stone space of complete $n$-types relative to a theory $T$ is $S_n(T)$. The $S$ stands for Stone space. Often, we want to consider  types with parameters coming from a subset $A$ of some model $M$ (types over $A$). Then we write $S_n(A)$ as shorthand for  $S_n(\text{Th}(M_A))$. This is the same as what you denote by $\text{Typ}_{M_A}(n)$.

Another common notational convention in model theory is to use single letters to denote tuples by default. So e.g. $p(x) = \text{tp}(a/B)$ could be an element of $S_n(B)$ if $x = (x_1,\dots,x_n)$ and $a = (a_1,\dots,a_n)$. This convention is not universal, but really de-clutters notation - and in many situations, there's really no meaningful difference between singletons and tuples.

2. Ok, now let's look at the statement (rewritten the way I would state it):

Let $A\subseteq B$ be sets, and let $p(x)$ be a partial type over $A$. Suppose that for any realization $a$ of $p(x)$, $\text{tp}(a/B)$ is isolated in $S_n(B)$. Show that $p(x)$ is equivalent to a formula over $A$.

You're right that this statement "takes place in the monster model". Here's the same statement, rewritten to make the monster model $\mathbb{M}$ explicit:

Let $A\subseteq B$ be subsets of $\mathbb{M}$, and let $p(x)$ be a partial type over $A$. Suppose that for any realization $a\in \mathbb{M}$ of $p(x)$, $\text{tp}(a/B)$ is isolated in $S_n(B)$. Show that $p(x)$ is equivalent to a formula over $A$.

If we drop the monster model convention entirely, here's the correct translation of the statement:

Let $A\subseteq B$ be subsets of a model $M$, and let $p(x)$ be a partial type over $A$. Suppose that for any realization $a\in N$ of $p(x)$ in an elementary extension $M\preceq N$, $\text{tp}(a/B)$ is isolated in $S_n(B)$. Show that $p(x)$ is equivalent to a formula over $A$.

Note in particular that $A$ and $B$ are just sets of parameters, not models. The  advantage of the monster model convention here is just that we don't have to quantify over models and elementary extensions each time we have an alternation of quantifiers in the statement. We never have to move to an elementary extension, because everything you would find in an elementary extension is "already there" in the monster.

3. Your topological proof of the statement is correct. Note that you're implicitly using two facts in the proof: (1) every clopen set in the Stone space $S_n(A)$ is definable by a formula over $A$ (this is why it suffices to show that the closed set defined by $p(x)$ is also open), and  (2) every surjective continuous map from a compact space to a Hausdorff space is a quotient map. These roughly correspond to the two main steps in the "concrete" proof I gave in the answer to the linked question: (1) The use of the compactness theorem to show that $p(x)$ is equivalent to a finite disjunction of formulas over $B$, and (2) the fact I cited that if a set definable by a formula over $B$ is invariant over $A$ then the set is actually definable by a formula over $A$.

4. As for "what is the significance of this property?" ... Well, to be honest, the statement in question doesn't actually seem that interesting or useful to me. One could view it as a generalization of the basic (and important) fact that if a partial type $p(x)$ is algebraic (has only finitely many realizations), then it is equivalent to a formula. [List the realizations of $p(x)$ as $b_1,\dots,b_k$ and let $B = A\cup \{b_1,\dots,b_k\}$. Then every realization of $p(x)$ has a complete type over $B$ which is isolated by one of the formulas $x = b_i$. Applying our property, we can conclude that $p(x)$ is equivalent to a formula over $A$.] Certainly isolated types are very important in model theory, and I could also imagine our property coming up as a step in a proof about atomic or prime models, or about Morley rank. If nothing else, it's a good exercise to get some practice thinking about Stone spaces.


答:
Here's a proof that shows how the condition on the 3rd derivative could be utilized.

Take Taylor's remainder theorem to 3rd order and apply it to the intervals $(x,x+h)$, $(x,x+k)$, $h\neq k$. Now consider only the first interval since the second one can be done similarly.

There exists $\xi\in(x,x+h)$ such that
$$f(x+h)=f(x)+f'(x)h+\frac{f''(x)}{2!}h^2+\frac{f'''(\xi)}{3!}h^3$$

Now since $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f(x+h)=a$, and $\lim_{x\to\infty}f'''(\xi(x,h))=0$ (why?) we conclude that
$$\lim_{x\to\infty}\left(f(x+h)-f(x)-\frac{f'''(\xi)}{3!}h^3\right)=h\lim_{x\to\infty}\left(f'(x)+\frac{h}{2}f''(x)\right)=0 ~~~~~~~~~~(1)$$
Similarly we conclude
$$\lim_{x\to\infty}\left(f'(x)+\frac{k}{2}f''(x)\right)=0~~~~~~~~~~~~~(2) $$

Taking the linear combinations $(2)-(1)$ and $h(2)-k(1)$ and exploiting the linearity of the limit operation when two limits exist, we obtain the desired result. The result is generalizable to a $C^n$ function whose $n$th derivative goes to zero for any $n$.

Body:
Given $f \in C^3(\mathbb{R}).$ Suppose that
$\lim\limits_{x\to+\infty}f(x)=a \in \mathbb{R}$ and
$\lim\limits_{x\to+\infty}f'''(x)=0$. Prove that
$\lim\limits_{x\to+\infty}f'(x)=\lim\limits_{x\to+\infty}f''(x)=0$.

My solution: Since $\lim\limits_{x\to+\infty}f(x)=a$, there exists $M>0$ that for all $x, x' \in \mathbb{R}$, if $x,x'>M$ then $f(x), f(x') \in \left(a-\varepsilon^2/2, a+\varepsilon^2/2\right)$. This yields
$$\vert f(x)-f(x') \vert < \varepsilon^2.$$

Therefore $\frac{\vert f(x)-f(x') \vert}{\varepsilon} < \varepsilon$. So, for all $x>M$, $x'=x+\varepsilon$ we have
$$0\le\frac{\vert f(x+\varepsilon)-f(x) \vert}{\varepsilon} < \varepsilon.$$
Let $\varepsilon \to 0$ we have
$$0 \le f'(x) \le 0 \text{ or } f'(x)=0.$$

That means $\lim\limits_{x\to+\infty}f'(x)=0$.

Doing similarly as above we have
$$\lim\limits_{x\to+\infty}f''(x)=0.$$

My question: $\lim\limits_{x\to+\infty}f'''(x)=0$ is really necessary or not (because I don't use it in my proof)?


问:
Interesting integral in changing integration bounds - sum
$$ \sum_{k=1}^{\infty} \int_{\frac{1}{k^3}}^{\frac{1}{k^2}} e^{- \left ( x^2 + |x| \right) } ~~ dx$$

How do I even begin calculating this horrifying thing?! I mean, I get the idea of the bounds being smaller and smaller thus making sense of this sum.. first I thought to myself that maybe we just need to find the integral itself, plug the bounds $\frac{1}{k^2}$ and $\frac{1}{k^3}$ and then take the sum (running on the variable $k$ of course) - but then what? It seems impossible! Is there another way of doing it?

This seems like a really hard problem for me, I would appreciate your help. Thanks!

答:
Not an answer, just too long for comments. Well, we can at least reduce it a bit using the normal distribution...

\begin{align*}
\sum_{k \ge 1 }\int_{1/k^3}^{1/k^2}e^{-(x^2+|x|)} dx = & \sum_{k \ge 1 }\int_{1/k^3}^{1/k^2}e^{-(x^2+x)} dx
 = e^{1/4} \sum_{k \ge 1 }\int_{1/k^3}^{1/k^2}e^{-(x-1/2)^2} dx\\
  = &  e^{1/4} \sqrt{\pi} \sum_{k \ge 1} (\Phi(1/k^2)-\Phi(1/k^3))
\end{align*}

where $\Phi$ is the normal distribution with mean $\frac 12$ and standard deviation $\frac{1}{\sqrt 2}$.