问:
Rudin RCA Problem 6.4
I'm trying to solve the following Problem 6.4 from Rudin's RCA:

Suppose that $1\leq p\leq \infty$, and $q$ is the exponent conjugate to $p$. Suppose that $\mu$ is a $\sigma$-finite measure and $g$ is a measurable function such that $fg\in L^1(\mu)$ for every $f\in L^p(\mu)$. Prove that then $g\in L^q(\mu)$.

I can get to the point where $\int_{E}|g|d\mu<\infty$ for every finite measure set $E$, but I'm not sure where to proceed. I would appreciate any suggestions.

Thanks!

答:
Here are a few hints: you can define $$X_k = \bigg\{ f \in L^p(\mu) : \|fg\|_1 \le k \|f\|_p \bigg\}.$$ By hypothesis $\cup_{k=1}^\infty X_k = L^p(\mu)$.

Show that each $X_k$ is closed. Then, according to the Baire Category Theorem, at least one $X_k$ has a nonempty interior. This means there exist $k \in \mathbf N$,  $f_0 \in L^p(\mu)$, and $\epsilon > 0$ with the property that if $f \in L^p(\mu)$ and $\|f - f_0\| \le \epsilon$, then $f \in X_k$.

If $f \in L^p(\mu)$ and $\|f\|_p \le 1$, then $\|(\epsilon f + f_0) - f_0\|_p \le \epsilon$, so that $\|(\epsilon f + f_0) g\|_1 \le k \|\epsilon f + f_0\|_p$. Use this fact to find a constant $C$ (depending on $\epsilon$, $k$, and $f_0$) such that
$$\sup_{\|f\|_p \le 1} \|fg\|_1 \le C.$$

Since the underlying space is $\sigma$-finite this gives you what you need.

Title: Rudin RCA Problem 6.4


问:
Homology of Identifications of 2-torus
While working through old algebraic topology quals, I found these questions and wasn't sure if I had the right idea.

Let $X$ be a closed genus–2 surface. Let $\alpha$ be a non-separating circle in $X$ and $\beta$ be a separating circle that is disjoint from $\alpha$ and not null–homotopic. Let $Y$ be the space obtained from $X$ by identifying $\alpha$ and $\beta$ by some choice of homeomorphism. Let $Z$ be the space obtained from gluing together two copies of $X$ along a homeomorphism of $\alpha$ and $\beta$. Compute the fundamental groups and integral homology of these spaces.

I've tried drawing the CW-complex structure of these spaces:

CW-Complex for X,Y,Z

My questions are:

Is this the right CW-complex structure?

How can I read the fundamental group/homology off of these? I'm confused in $Y$ if there is still only one 2-cell with the way $\alpha$ is drawn, and how I would read the fundamental group in either case with the way $\alpha$ now cuts the CW complex.

Edit: Following Kevin's answer, I have two possible CW complexes? I'm unsure whether the separating circle in the larger copy of X gets identified with the non-separating circle $\alpha$.

Option 1

Option 2

Edit 2: Option 1 it is!

答:
The images of $X$ and $Y$ seem fine to me.

For your second question(s), $Y$ certainly continas two $2$-cells, where $e_1^2$ is glued via the attaching map $zyz^{-1}y^{-1}\alpha^{-1}$ and $e_2^2$ is glued via $\alpha x\alpha x^{-1}\alpha^{-1}$. So you can obtain a group presentation $$\pi_1(Y)=\langle x,y,z,\alpha\mid zyz^{-1}y^{-1}\alpha^{-1}, \alpha x\alpha x^{-1}\alpha^{-1}\rangle$$
This result follows from Proposition 1.26 of Hatcher.

The image of $Z$ is confusing for some reason:

The question asks you to glue $\alpha,\beta$ of two copies of $X$ but not other identified edges, so you should really change the letters of the labelling  scheme $yz...$ of one copy. Otherwise it's like $z$'s of the two copies are also identified...

Once these corrections are done, the CW structure is again clear, and you can compute the fundamental group by the same technique.

Hints for computing homology groups:

The cell structure yields a cellular chain complex $$0\hookrightarrow\Bbb{Z}^m\overset{\partial}{\to}\Bbb{Z}^n\twoheadrightarrow0$$ where $m,n\in\Bbb{N}$.

$\partial(e_i^2)$ is a linear combination of $e_j^1$'s, which form the 'boundary'. This is exactly what the attaching map tells you.

You may try solving the problem yourself before telling me to add the full answer.

ContentLicense: CC BY-SA 4.0

Title: Homology of Identifications of 2-torus


问:
What do we call an exponential object in a concrete category that basically equals the one in Set
So suppose you've got a concrete category $\mathcal C$ and two objects $X, Y$ of $\mathcal C$. For instance, $\mathcal C$ could be the category of topological spaces and we could also suppose that $Y$ is locally compact, so that $X^Y$, if endowed with the compact-open topology, is an exponential object of $X$ to the power of $Y$.

Now in this scenario, the space $X^Y$ is nothing but a subset of the set $X^Y$ (that is, the continuous functions from $Y$ to $X$) with a topology on it, so that the underlying set happens to be a subset of the exponential object in the category $\mathbf{Set}$. Moreover, all the functions associated to this exponential object are merely restrictions of those given in the category $\mathbf{Set}$.

Now my question is:

Is there a special name for an exponential object of $X$ to the power of $Y$ which, when projected to $\mathbf{Set}$, is nothing but a subset of the set $X^Y$, together with the canonical evaluation map and the canonical assignment of the $\lambda$ function as explained in the Wikipedia article "Exponential object"?

答:
The example of $\text{Top}$ has the additional property that the underlying set functor is represented by the terminal object $1$; in other words, it is the functor $\text{Hom}(1, -)$ of global points. This is not true of many concrete categories, but when it is true, and assuming as usual that "concrete category" means the underlying set functor is faithful, applying $\text{Hom}(1, -)$ to an evaluation homomorphism

$$X \times Y^X \to Y$$

produces the concrete evaluation homomorphism

$$\text{Hom}(1, X) \times \text{Hom}(X, Y) \to \text{Hom}(1, Y)$$

given by the action of $\text{Hom}(X, Y)$ on global points. Currying this evaluation homomorphism gives a map $\text{Hom}(X, Y) \to \text{Hom}(1, Y)^{\text{Hom}(1, X)}$ and you want to know when this map is an injection. This is guaranteed by the assumption that $\text{Hom}(1, -)$ is faithful, which says exactly that a morphism is determined by its action on global points.

So the key condition is that the underlying set functor is (faithful and) the global points functor; we might call such a category globally concrete (I don't know if it has a standard name).


问:
I solved the problem. But I have some doubts on the solution.
Given the sequence $a_n = \sin(n)$, where $n$ is an integer. Prove that limit superior is $1$.

My solution: by previous theorems we know that if $x,y \in \Bbb R$ and the ratio of $x$ and y is irrational, then there exists the integers $a$ and $b$ such that $$\forall \epsilon, \quad  0<ax+by<\epsilon$$ So we can put integrs $m$ ands $n$ such that $$0<\frac{n}{m}-\frac{\pi}{2}<\epsilon$$ And we are done.

But here m should be in form of $m=4k+1,\, k\in \Bbb Z$.

My question is, how we can be sure that we can find such $m$?

答:
Let $x$ and $y$ be real numbers. And suppose that $ \frac{x}{y}$ is irrational.

Prove the following: for each $\varepsilon>0$, there exist integers a and b such that $0<ax+by< \varepsilon$.

Consider the set $S=${$ax+by>0$: $x,y \in R$, $ \frac{x}{y}$ is irrational, $a,b \in Z$}.

First to show $d= \inf S \ge 0$ exists we must show $S$ is non-empty and $0$ is lower bound of $S$. The set $S$ is obviously non-empty. Just consider $|x|+|y|=±1⋅x+(±1)⋅y>0$. And by definition, if $ax+by∈S$ then $ax+by>0$ so $0$ is a lower bound.

Assume that $c=min$ $S$. Then I claim that $x$ and $y$ are both integer multiples of $c$.
$x=cm$, $y=nm$ and $n,m \in Z$.

If not then $x=cm+r$ where $0\leq r<c$ and $ m \in Z$. From here we have
$r=x - mc= x - m(kx+ty)=(1-mk)x+(-tm) y$ and $k,t \in Z$. So $r \in S$ and $r<c$. It contradicts with the fact that $c =\min S$. Similarly we can show $y=cn$ for some $n \in Z$.

But $ \frac{x}{y} = \frac{m}{n}$ sould be irrational where $ m,n \in Z$ which is immposible. It means that the set $S$ has no minimum.

Now..... Suppose $d=\inf S≠0$. Then $\inf S>0$. That mean $\frac{3}{2}d>d$ so $2d>d$ is not a lower bound of $S$. We also know $d∈S$ is impossible as that would mean $d=\min S$ and we know $\min S$ doesn't exist.

So there exists $ax+by$ so that $d<ax+by<2d$. And as $ax+by>d$, $ax+by$ can't be a lower bound so there must exist $ex+fy$ so that $d<ex+fy<ax+by<2d$.

So $0<(a−e)x+(b−f)y<d=\inf S$ which is impossible as $0<(a−e)x+(b−f)y⟹(a−e)x+(b−f)y∈S$.

So $d=infS=0$.

.......

Which means for any $ϵ>0$, then $ϵ$ is not a lower bound of $S$ so there is an $ax+by∈S$ so that $0<ax+by<ϵ$.


问:
Rank of the derivative matrix.
I was working in some exercises about multivariable calculus and I'm stuck. The exercise is

Lef $f:U\subseteq\mathbb{R}^{n}\to\mathbb{R}^{m}$ be a $C^1$ function where $U$ is an open set and $n<m$. If $x_0\in U$ and $\operatorname{rank}(Df(x_0))=n$ (rank of the matrix derivative) then there exist $A\subseteq U$ an open set such that $x_0\in A$ and $\operatorname{int}(f[A])=\emptyset$ (topological interior).

I really don't know how to procced but I have an idea: as $n<m$ and $\operatorname{rank}(Df(x_0))=n$ then $Df(x_0)$ has a submatrix $B$ of size $n\times n$ such that has non zero determinant. Suposse that $f=(f_1,\dots,f_m)$ where $f_{i}:U\subseteq\mathbb{R}^{n}\to\mathbb{R}$ and define $h:U\subseteq\mathbb{R}^{n}\to\mathbb{R}^{n}$ by $h(x)=(f_1(x),\dots,f_n(x))$ (w.l.g. the first $n$ rows of $Df(x_0)$ are l.i.). Then, by the inverse function theorem, there exist a neighborhood $V$ of $x_0$ where $h$ is invertible. But, from here, how can I go? Thanks for your help.

答: Indeed, $f(V)$ has empty interior under your assumptions. To see this, take $x \in V$ and write $y=(y_1,\ldots,y_m)=f(x)$. Observe that $y_{n+1}, \ldots, y_m$ are determined by $y_1,\ldots,y_n$ since $y=f(h^{-1}(y_1,\ldots,y_n))$. Thus for any $\epsilon \ne 0$ the point $(y_1,\ldots, y_{m-1},y_m+\epsilon)$ is not in $f(V)$.

Title: Rank of the derivative matrix.


问:
Quick Area of Cross Section Question
"The base of a solid is the region in the first quadrant between the graph of $y=2x$ and the $x$-axis for $0\leq1\leq2$. For the solid, each cross section perpendicular to the $x$-axis is a quarter circle with the corresponding circle's center on the $x$-axis and one radius in the $xy$-plane. What is the volume of the solid?"

I guess I'm a bit confused about both the wording and where I'd have to go in this question, I understand how to find the area of a cross-section when these cross-sections might be a square or a rectangle, but I'm a bit lost when it comes to circles as well. Any help would be appreciated!

答:
THe method of using integraion to find find volume that you are using is

Pick an axis.

Slice the volume in slices perpendicular to that axis.

Integrate the volume of the slices with respect to the coordinate along that axis.

(Generally, the axis need not be a coordinate axis, but it is convenient when this happens.)

When the problem has

For the solid, each cross section perpendicular to the $x$-axis is a quarter circle with the corresponding circle's center on the x-axis and one radius in the $xy$-plane.

You are told that the slice is a quarter circle, so has one quarter the area of a circle, and that (1) all the radii for that quarter circular are perpendicular to the $x$-axis and (2) one radius is in the $xy$-plane.  This means that in a slice, one can take the distance perpendicular to the $x$-axis to the line $y = 2x$ as the radius of the circle, one quarter of which area is accumulated into the integral.  (In the $xy$-plane, a perpendicular to the $x$-axis is parallel to the $y$-axis.  A ray starting on the $x$ axis, in the first quadrant of the $xy$-plane, and proceeding perpendicular to the axis crosses the triangle base of the volume, reaching the end of the base at the line $y = 2x$.  (If this is not clear, sketch a graph of the base.)

Suppose $r$ is the radius of a circle.  The area of that circle is $\pi r^2$.  The area of one-quarter of that circle is $\frac{1}{4} \pi r^2$.  The distance from the $x$-axis to the line $y = 2x$ (the radius described by the ray in the previous paragraph) is $2x$, so the radius we are to take for the slice at coordinate $x$ is $r = 2x$.  Then the area contribution from the slice at coordinate $x$ is $\frac{1}{4} \pi (2x)^2$.  That is the integrand that we integrate from $x = 0$ to $x = 2$.  (Again, if it is not clear that this is the range of coordinate that includes all the slices, sketch the base.)

The following diagram of the volume is from a point in the $(+++)$-octant looking back throught the volume at the origin.

Example slices (for $x \in \{1/2, 1, 3/2, 2\}$ are shown as fairly clear quarter circles.  The base is the $xy$-plane, at the bottom of the diagram.  The dashed gray line is in the $xz$-plane and marks the ends of the radii of the quarter circles in that plane.

We can more clearly see the "slicing" viewing from the $(+-+)$-octant nearly parallel to the positive $y$-axis.

This angle reinforces that we are given the axis, we are given the area formula for the slices, and all we have to find is the relation between the $x$-coordinate and the area of the corresponding slice.

Title: Quick Area of Cross Section Question


问:
How do i express a in terms of the other vectors?
I've been having issues solving this math problem. I just can't seem to understand how to incorporate the two c vectors. I've tried 2c - b and -b + c +c (both equivalent) and I've gotten it incorrect. Is there someone who can possibly lead me to the right path? Thanks!

Triangle made up of vectors a, b, and c

答:
Observe from the picture that:

$$2\vec{c} + \vec{a} = \vec{b}$$

Now solve for $\vec{a}$:

$$\vec{a} = \vec{b} - 2\vec{c}$$


问:
Question on the 2020 RMO.
In the “expression”
each of the twelve @ symbols is replaced with either $\times$ or $÷$ such that the value of the resulting
expression is an integer.

1@2@3@4@5@6@7@8@9@10@11@12@13

Find the greatest common factor of all such integer values.

答:
First every such integer is divisible by $7, 11, 13$.

Highest power of $2$ of $13!$ is even, so you need to find an integer product such that it's odd. This can be easily done.

Then the highest power of $3$ of $13!$ is odd, so you need to find an integer product such that it's divisible by $3$ but not by $9$. This is easy too.

The highest power of $5$ of $13!$ is $2$. It's easy to find one integer product that's not divisible by $5$.

Therefore the $\gcd$ of all integer products is $3\times 7 \times 11 \times 13 = 3003$.

Title: Question on the 2020 RMO.


问:
Solving indetermination in limit
I'm trying to solve this limit, but I can't get it out of a $\frac{0}{0}$ indetermination:

$$\displaystyle \lim_{x \to 4} \; \frac{x-4}{5-\sqrt{x^2+9}}$$

Maybe there is something I'm missing. Thanks a lot in advanced.

答:
For all $x\ne 4$ one has
$${x-4 \over 5-\sqrt{x^2+9}}={(x-4)(5+\sqrt{x^2+9}) \over (5-\sqrt{x^2+9})(5+\sqrt{x^2+9})}={(x-4)(5+\sqrt{x^2+9})\over 16-x^2}=-{5+\sqrt{x^2+9}\over x+4}\ .$$
Here the right side is continuous at $x=4$ and has the value $-{5\over 4}$ there. This means that the limit you want is $-{5\over 4}$.</p>

ContentLicense: CC BY-SA 3.0

Title: Solving indetermination in limit


问:
Given the joint density function, are the random variables independent or dependent?
Question: Let a joint density function $f_{X,Y}: \mathbb{R}^2 \rightarrow \mathbb{R}$ be given by:

\begin{equation}
f_{X,Y}(x,y) =
\begin{cases}
cx^2y \quad &\text{if }0<x<1 \text{ and max}\{b,x\} <y <2\\
0 &\text{else}
\end{cases}
\end{equation}
where $b\in[0,1]$.

We take for granted that for every $b$, there exists some constant $c \in \mathbb{R}$ such that $f_{X,Y}$ is a valid joint density function.

Are random variables X and Y independent?

My solution: So the solution was independent for $b = 1$, and dependent for $b < 1$. I only managed to solve it for $b = 1$.

I solved it by finding the value $c$ which makes the above a valid density (i.e. integral over $\mathbb{R}^2 = 1$), computing the marginals $f_X$ and $f_Y$, and showing that $f_{X,Y} = f_X f_Y$ which is true iff X and Y are independent.

This was simple to compute because the bounds for integration were clear. i.e. $0<x<1, \quad 1<y<2$.

However, I'm not sure how to solve it for $b < 1$. I attempted to use the same approach of computing the marginals, but I couldn't figure out a way to make the bounds of integration work.

i.e. I wasn't sure how to compute:

$f_X = \int^{\infty}_{-\infty} f_{X,Y} dy, \quad f_Y = \int^{\infty}_{-\infty} f_{X,Y} dx, \quad$ given the bounds  $0<x<1 \text{ and max}\{b,x\} <y <2$.

Is there a clean way to compute the above integrals? My multivariable calculus is a little rusty. Thank you!

答:
Given that there is a joint density, the variables are independent iff this joint density can be written as a product: $f_{XY}=h(x)g(y)$ for a.e. $x,y \quad$(*)

Indeed, given (*) you can easily deduce that $h=f_X$ and $g=f_Y$.

This settles the case $b=1$.

For $b<1$, check that both events $\{X>(b+1)/2\}$ and $\{Y<(b+1)/2\}$ have positive probability, yet their intersection has probability $0$.


问:
How to find the period of this sinusoid?
I'm stuck trying to find the period of this sinusoid and would really like some pointers to different ways to approach this problem.

$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3})$$

I would like to turn the product into a sum and do something about that $2$ in the exponent of $\sin$

$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3}) = \cos(\frac{4\pi t}{5})\left[ 1 - \cos(\frac{16 \pi t}{3}) \right]\frac{1}{2} $$

$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \cos(\frac{4\pi t}{5}) \cos(\frac{16 \pi t}{3})\right]$$
$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \frac{1}{2}\cos(\frac{4\pi t}{5} + \frac{16 \pi t}{3}) \cos(\frac{16 \pi t}{3} - \frac{4\pi t}{5})\right]$$

So the sum of cosines will have a period that is the LCM of of the periods of the individuals terms we added. The angular frequencies are $\frac{4\pi}{5}, \frac{92\pi}{15}, \frac{68\pi}{15}$.

Converting this to periods -> we get $\frac{5}{2}, \frac{15}{46}, \frac{15}{34}$, but the answer is $\frac{30}{4}$

答:
To get the LCM, we want the smallest $T$ which is an integer multiple of the three periods. That is, we want integers $n_1,n_2,n_3$ such that $$T=\frac{5}{2}n_1=\frac{15}{46}n_2=\frac{15}{34}n_3$$ with $T$ as small as possible.

To this end we first focus on the middle equality, which rearranges to $23n_1=3n_2$. Since $23$ and $3$ are prime, this requires that $23$ divides $n_2$ and $3$ divides $n_1$. Hence there must be an integer $k$ such that $n_1=3k$ and $n_2=23k$.

With this, the third quantity of the period equality becomes redundant and we have $$T=\frac{15}{2}k=\frac{15}{34}n_3.$$

From this we deduce the value of $n_3$ and therefore know $T$ in terms of $k$. From this we deduce the smallest such value and conclude. (I leave the reader to fill in the details.)


问:
Check if two lines intersect
I have a non-linear line with these coordinates:

```<code>(1,50)(2,40)(3,45)(4,40)(5,60)(6,30)
</code>```

When I draw the line passing through each of these points, noted line A, I can visually see that another straight line drawn from last to first point only - from (1,50) to (6,30) - noted line B, would intersect/cross line A.

My goal is to be able to mathematically check that line B does NOT intersects with line A, is there some equation or approach I could go with to check that?

EDIT:
Sorry for the misleading "line" word I use, I should have used curve. Here is the representation of what I meant (credits to Joffan):

答:
Got an elegant answer here, with two functions (in python, but Mathematically readable):

```<code>def ccw(A,B,C):
    return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)

# Return true if line segments AB and CD intersect
def intersect(A,B,C,D):
    return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)
</code>```

Joffan helped me reformulate my problem with his comment I suspect that your "line A" is a curve made of linear segments which made my research much easier with a lot more results like the mentioned solution.

Title: Check if two lines intersect

Body:
Got an elegant answer here, with two functions (in python, but Mathematically readable):

def ccw(A,B,C):
    return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)

# Return true if line segments AB and CD intersect
def intersect(A,B,C,D):
    return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)

Joffan helped me reformulate my problem with his comment I suspect that your "line A" is a curve made of linear segments which made my research much easier with a lot more results like the mentioned solution.


问:
A visual interpretation of probability?
I was using the following website for a visual information theory : http://colah.github.io/posts/2015-09-Visual-Information/.

And I came across the following:

So it says that in the first image where the lines are straight the events are independent.  But in the second picture when the events are not independent then the lines are bloated. I am wondering why the reason is so. It says here that some of the areas are bloated and some are not, and this is how we know that one event is more probable than the other. But how does that factor into independence? Even in the first picture, one portion of the area is bigger than the other. So how does the area matter or give us an answer about independence?

答:
While I think that that visual is not particularly helpful, in no small part because the exterior is nonsensical, there is a way to interpret it reasonably. In the first case, you can pick any information from the clothing axis, and gain no more information about the weather. Similarly, you can pick any information from the weather axis and gain no more information about the clothing. This is to say that if we compare the whole graph to a subset obtained by focusing on part of one axis, the graph is largely unchanged.

On the other hand, this process fails in the second diagram. If I restrict to "coat" section, the graph is telling me that it is more likely to be raining - I have gained more knowledge about the weather variable by being more specific in my investigation of the clothing variable.

This is the crux of independence and dependence. Two variables are independent if adding information about one does not give you new information about the other.

Title: A visual interpretation of probability?


问:
Differentiability of $ f(x,y)=\frac{2x^3+3y^3}{x^2+y^2}$
Let $f:\Bbb R^2 \rightarrow \Bbb R $ be defined by
$$f(x)= \begin{cases}
       f(x,y)=\dfrac{2x^3+3y^3}{x^2+y^2}\ \ \ &(x,y)\ne(0,0)   \\
      0 & (x,y) = (0,0) \\
   \end{cases}
$$

Let $f_x(0,0)$ and $f_y(0,0)$ denote the first order partial derivative of $f(x,y)$  with respect to $x$ and $y$ respectively,at the point $(0,0)$. The which one of the following statements is TRUE?

$(A)$ $f$ is continuous at $(0,0)$ but $f_x(0,0)$ and $f_y(0,0)$ do not exist.

$(B)$ $f$ is differentiable at $(0,0)$

$(C)$ $f$ is not differentiable at $(0,0)$

$(D)$ $f$ is not continuous at $(0,0)$ but $f_x(0,0)$ and $f_y(0,0)$ exist.

My attempt:

I was checking continuity so:

Choosing path $y=mx$

$\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3y^3}{x^2+y^2}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3m^3x^3}{x^2+m^2x^2}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3m^3x^3}{x^2+m^2x^2}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{x(2+3m^3)}{1+m^2}=0$

$f(0,0)=0=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3y^3}{x^2+y^2}$

Makes it continuous so option $D$ cancels out

Checking partial differentiability:

$f_x(0,0)=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^3+3y^3}{x^2+y^2}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{(x^2+y^2)\cdot(6x^2)-(2x^3+3y^3) \cdot (4x)}{(x^2+y^2)^2}$

Choosing path $y=mx$

$\lim_{(x,y)\rightarrow(0,0)}\dfrac{(x^2+m^2x^2)\cdot(6x^2)-(2x^3+3m^3x^3) \cdot (4x)}{(x^2+m^2x^2)^2}$

$\lim_{(x,y)\rightarrow(0,0)}\dfrac{(1+m^2)\cdot(6)-(2+3m^3) \cdot (4)}{(1+m^2)^2}$

For different values of $m$ we will have different answers so the limit is not unique.

$A$ option cancels out

So a function to be differentiable it must satisfy the following:

$$\lim_{(h,k)\rightarrow(0,0)}=\dfrac{f((0+h),(0+k))-f(x,y)-h \cdot f_x(0,0)-k \cdot f_y(0,0)))}{\sqrt{h^2+k^2}}$$

but $f_x$ and $f_y$ doesn't exist so its not differentiable.

Am I thinking it right so only option left is $C$

答:
For the first part we have that

$$\frac{|2x^3+3y^3|}{x^2+y^2} \leq \frac{2|x^3|}{x^2+y^2}+\frac{3|y^3|}{x^2+y^2} \leq 2|x|+3|y|$$

thus the limit goes to $0$ by squeeze theorem and the function is continuous. For differentiability we must compute the partials at $(0,0)$ by definition:

$$f_x(0,0) = \lim_{x\to0} \frac{f(x,0)-f(0,0)}{x} = \lim_{x\to0} \frac{2x-0}{x} = 2$$

and similarly $f_y(0,0) = 3$. Therefore we need to check whether the following limit exists and equals $0$:

$$\lim_{(x,y)\to0} \frac{f(x,y)-f(0,0)-2x-3y}{\sqrt{x^2+y^2}}$$

I'll leave this last step for you to check.


问:
Difference of two powers of $3$ divisible by $2011$
How to prove that there exists two powers of $3$ that differ by a number that is divisible by $2011$?</p>

答: Imagine you calculate $3^n \pmod {2011}$ for a set of $2012 n$'s.  Two of these must match.</p>


问:
Is every direct product of (finite) cyclic groups abelian?
Long story short I'm halfway through a proof and have hit a step where I show that a group order $p^2$ is abelian. I did this by splitting into the $C_{p^2}$ case and the $C_p\times C_p$ case, the first being trivial as all cyclic groups are abelian. The second case I proved by using the definition of the direct product and showing it directly from the definition of the two $C_p$ groups.

However this feels odd. Something feels off - especially added to the fact that my proof seemingly holds for any two cyclic groups, implying the direct product of any two cyclic groups is abelian, and yet I can't find anything online talking about this. Is it just so trivial nobody mentions it? The only thing I consistently see brought up is that "every cyclic group is an abelian group" (trivial), and "every finitely generated abelian group is a direct product of cyclic groups", which is the converse of what I'm trying to show.

答:
If you look at the way the direct product is defined, in particular the group operation, it's pretty easy to see that the answer is yes.  For we have $(a,b)*(c,d)=(ac,bd)=(ca,db)=(c,d)*(a,b)$.

Thus every direct product of abelian groups is abelian.


问:
The floored quadratic equation $a \lfloor x^2 \rfloor + b \lfloor x \rfloor + c = 0$
What are the roots of $f(x) = a\lfloor x^2 \rfloor +b\lfloor x \rfloor + c$?
(where $a \neq 0$ and $a,b,c \in \mathbb{R}, f: \mathbb{R} \rightarrow \mathbb{R}$ as usual)

My failed approach:
Let $\delta = x - \lfloor x \rfloor$ and get
\begin{align}
a \lfloor (\lfloor x \rfloor + \delta)^2 \rfloor + b\lfloor x \rfloor + c &= 0, \\
a \lfloor \lfloor x \rfloor^2 + 2\lfloor x \rfloor \delta + \delta^2 \rfloor + b\lfloor x \rfloor + c &= 0, \\
a \lfloor x \rfloor^2 + a\lfloor 2\lfloor x \rfloor \delta + \delta^2 \rfloor + b\lfloor x \rfloor + c &= 0.
\end{align}

From here there are 2 cases:
$$ a \lfloor x \rfloor^2 + a\lfloor 2\lfloor x \rfloor \delta \rfloor + b\lfloor x \rfloor + c = 0 \ \text{or}\ a \lfloor x^2 \rfloor + a\lfloor 2\lfloor x \rfloor \delta \rfloor + b\lfloor x \rfloor + (c + 1) = 0,$$

and then I'm stuck from here. This approach is probably not helpful since even if you get a quadratic equation in terms of $\lfloor x \rfloor$, you still have $\delta$ which is in terms of $x$, the variable of $f(x)$. Which is a difficult.

答:
Fix $a,b,c$ reals with $a\neq 0$, let us look at some necessary conditions that $$(\dagger)\quad\quad a \lfloor x^2\rfloor + b \lfloor x\rfloor +c =0$$ will have a root:

(1) It is necessary that $a,b,c$ satisfies an integral relation $$aN + b M +c =0$$ for some integers $N,M$. Otherwise we will never have a root. For instance if $a,b$ are both rational and $c$ is irrational, then there is no root to $(\dagger)$. This is a very weak necessary condition to meet. We know that $N$ needs to be roughly the square of $M$, so we will make this precise next on what integers $N,M$ can be.

(2) Suppose $(\dagger)$ has a solution for some $x$. Let's break it into cases.

If $x\ge 0 $, then $n\le x < n+1$ for some non-negative integer $n$. In this case $\lfloor x\rfloor =n$, while $n^2\le x^2 <(n+1)^2 = n^2 + 2n+1$. Hence $\lfloor x^2 \rfloor$ can take on any of $n^2, n^2+1 ,\ldots, n^2 +2n$. Hence for $(\dagger)$ to have non-negative $x$ solutions, we need to satisfy
$$
(\triangle)\quad\quad a (n^2 + k) + bn +c =0
$$
for some non-negative integer $n$ and $k=0,1,\ldots 2n$.

Once we have such $n$ and $k$, then we can work out what $x$ is.

We can analyze the scenario for a negative $x$ solution similarly.

If there is a negative $x$ solution, then $n\le x <n+1$ for some negative integer $n$. Note $(n+1)^2 < x^2 \le n^2$ as $x$ and $n$ are negative. So $\lfloor x \rfloor=n$ and $\lfloor n^2 \rfloor$ can take on any of $n^2 + 2n+1,\ldots ,n^2$ (again, note $n$ is negative).
Hence for $(\dagger)$ to have non-negative $x$ solutions, we need to satisfy
$$
(\triangle\triangle)\quad\quad a (n^2 - k) + bn +c =0
$$
for some negative integer $n$ and $k=0,1,\ldots -1-2n$.

Remark. Since we can actually solve for the real roots to $at^2+bt+c=0$, that will give you an approximate search range for such $n$, and you can brute force test out the relevant $(n,k)$ pairs that will satisfy $(\triangle)$ or $(\triangle\triangle)$. And then you can recover your $x$ values from $(n,k)$. (Note each $(n,k)$ pair will give you an interval of $x$ values.)


问:
Notation for multi-variate functions?
I have a question regarding the syntax for multivariate functions and corresponding definitions.  My understanding so far is a function accepts a single argument and this can (should?) be a multi-argument list passed as an ordered tuple.

If:

$f: X \rightarrow Y_0 \times Y_1$

$g: Y_0 \times Y_1 \rightarrow Z$

and an ordered tuple is the result of $f$ and an argument to $g$.

Commonly defined:

$(g \circ f)(x) = g(f(x))$

Should this definition really be:

$(g \circ f)x = gfx$

omitting the parenthesis because the result of $f$ is an ordered tuple?  I.e. already is enclosed in parenthesis.

Confusing the matter more, Wiki says that a

function of n variables is a function whose domain is a set of n-tuples

and says for a function:

$f : X_1 \times ... \times X_n \rightarrow Y$

When using function notation, one usually omits the parentheses surrounding tuples,

So when writing the form $f((x_1, ..., x_n))$, the parenthesis is omitted and instead we write $f(x_1, ..., x_n)$.

I'm a bit confused and could use guidance.  For example, I see Wiki define $(f + g)(x) = f(x) + g(x)$ and am asking myself, does this definition work with multi-variate functions.  Again, I would expect this definition to be $(f + g)x = fx + gx$ for a multi-variate argument.  I recall seeing some older mathematics literature and saw use with parenthesis omitted.  For this reason, I'm expanding to math-history.

答:
The parentheses for an element of a product space do not have the semantics of the parenthesis for evaluation of a function.  The way you are writing expressions suggests you interpret
$$  f(x)  $$
as "$f$" and "$(x)$", where the mere mention of $f$ means to use it for evaluation and the parenthesized expression is a $1$-tuple.  This is not the semantics of those uses of those symbols.  $f$ is the name/label of a function, the parentheses have the semantics of evaluation -- in particular, evaluate the function named prior to the parentheses on the input enclosed in the parentheses.

The expression $fx$ is not normally a valid sentence since it is two adjacent nouns.  If you intend to evaluate $f$ at $x$, parentheses are needed.

Consequently, in
$$  f((x_1, x_2, \dots, x_n))  $$
the outer parentheses have the semantics of function evaluation and the inner parentheses have the semantics of $n$-tuple aggregation.  It is a lexicographical hack (to improve legibility) to combine the doubled open parentheses into a single open parenthesis and the doubled close parentheses into a single close parenthesis.  This combination is merely one of presentation in textual form, it represents the doubled form, with both evaluation and aggregation parentheses.

Title: Notation for multi-variate functions?


问:
Side limits of ${\displaystyle \lim_{x\to0}\frac{|\sin\left(\frac{1}{x}\right)|}{1 + 2^{1/x}}}$
I need to find the side limits of this function if they exist. I've tried dividing both denominator and numerator with $\frac1x$, but that didn't help either. According to the book, the left-side limit doesn't exist, and the right-side limit is $1$. How could I solve this problem?

$${\displaystyle \lim_{x\to0}\frac{|\sin\left(\frac{1}{x}\right)|}{1 + 2^{1/x}}}$$

答:
\begin{align}
L &=\lim_{x\to 0} \frac{|\sin(1/x)|}{1+2^{1/x}}\\
&= \lim _{t\to \infty \:}\frac{\left|\sin \left(t\right)\right|}{1+2^t}\\
& \leq \lim _{t\to \infty \:}\frac{1}{1+2^t}\\
& = 0
\end{align}


问:
Is material implication consistent?
How do we know that material implication(and other definitions in logic) do not lead to contradictions? The only insight I have been able to find into this has to do with paradoxes involving material implication, which seems to have to do with differences between classical logic and ordinary language. Is there any way to intuitively understand its consistency?</p>

答:
The question is rather vague: What would it mean for a definition to lead to contradictions? Are you specifically concerned about material implication, or the basics of logic more generally? Please clarify if I've misinterpreted your question, but it sounds like you're concerned about the possibility that classical propositional logic (or a similarly basic logical system) proves a contradiction. That is, there is a sentence $\varphi$ such that $\varphi\land \lnot \varphi$ is provable in classical propositional logic.

You ask "how do we know" that this doesn't happen? Well, we give a proof. The whole idea of mathematical logic is to study logical systems as mathematical objects the same way we study natural numbers, groups, topological spaces, etc. And yes, we can easily prove that there are no contradictions in propositional logic.

Here's a sketch of how the proof goes. We say a sentence $\psi$ is valid if whenever we assign truth values to the basic proposition letters appearing in the sentence, the sentence comes out true (think about filling out a truth table). For example, $(P\land Q)\lor \lnot P \lor \lnot Q$ is a valid sentence. Now we proceed, by induction on the length of a proof, to show that every provable sentence is valid (this is called soundness). To carry out the inductive step, we assume that all the previous lines of the proof are valid sentences, and check that any way we could apply a proof rule to obtain the next line of the proof would produce a valid sentence. Since $\varphi\land \lnot \varphi$ is not valid (in fact, it always comes out false), this shows that $\varphi\land \lnot \varphi$ is not provable.

Of course, this is an ordinary mathematical proof, which happens, like all of mathematics, in a logical setting. If you are a true skeptic about things like the consistency of propositional logic, you may not find the above satisfying, because you do not accept any logical arguments. To make mathematics work, we have to have some bedrock agreement about what forms of argument we accept - and material implication is part of this bedrock. Deep skepticism about the bedrock, or a search for extralogical arguments for or against it, is probably better suited for philosophy.stackexchange.com.

Title: Is material implication consistent?


问:
Why doesn't integrating the area of the square give the volume of the cube?
I had a calculus course this semester in which I was taught that the integration of the area gives the size (volume):

$$V = \int\limits_a^b {A(x)dx}$$

But this doesn't seem to work with the square. Since the size of the area of the square is $x^2$ then $A(x) = {x^2}$, then:

$$V = \int\limits_{ - r}^r {{x^2}dx}  = \left[ {\frac{{{x^3}}}{3}} \right]_{ - r}^r = \frac{{{r^3}}}{3} - \frac{{ - {r^3}}}{3} = \frac{2}{3}{r^3}$$

It's clear that this is not the volume of the cube. Why is this the case? Am I misunderstanding something?

答:
Actually, you have two errors there:

The minor one is that you seem to want a cube of side $2r$,
since your integral goes from $-r$ to $r$.

The major error, as others have said, is that you are finding the volume of a pyramid, not a cube. Actually, since you are integrating from $-r$ to $r$,
you are finding the volume of two pyramids, one upside-down, touching at their points. That is why you get $2r^3/3$, when the volume of one pyramid is $r^3/3$.


问:
How come if $X$ is a Geometric random variable, then $P(X > x) = q^x?$
I know that $P(X > x) = 1 - P(X \leq x)$ via the definition of the complement, but how come this is also equal to $q^x$ where $q$ is the probability of failure in a trial in a Geometric distribution? How does $q^x$ account for all cases where $X > x$? Can someone explain this to me?

Thank You!

答: $X>x$ means that your first $x$ trials were all failures. The probability of this is clearly $q^x$.


答:
Let $X \sim $ Geometric$(p)$ then we know the CDF of $X$ is $F(k) = 1- (1-p)^k = 1-q^k$

So then what does $1- F(k) = P(X > k)$ mean then? It means at least $k$-many trials have failed.

Symbolically we have $P(X > k) = 1- P(X \le k) = 1 - [1 - (1-p)^k] = 1- [1-q^k] = q^k$

This is for the definition of Geometric distribution supported on $k \in \{1,2,3,4, \ldots\} = \mathbb{N}\backslash\{0\}$

Title: How come if $X$ is a Geometric random variable, then $P(X > x) = q^x?$

Body:
I know that $P(X > x) = 1 - P(X \leq x)$ via the definition of the complement, but how come this is also equal to $q^x$ where $q$ is the probability of failure in a trial in a Geometric distribution? How does $q^x$ account for all cases where $X > x$? Can someone explain this to me?

Thank You!


答:
Another equivalent way to say that

the first success to occur after $x$ trials,

is

you must failed at the first $x$ trials

What happens after $x$ doesn't matter.


问:
How few $(42^\circ,60^\circ,78^\circ)$ triangles can an equilateral triangle be divided into?
This is the parallel question to this other post with many answers already, in the sense that the $(42^\circ,60^\circ,78^\circ)$-similar triangles form the only non-trivial rational-angle tiling of the equilateral triangle (and the regular hexagon), modulo a real conjugation of the coordinate field (a subfield of $\mathbf{Q}(\zeta_{60})$) which transforms between $(42^\circ,60^\circ,78^\circ)$-similar triangles and $(6^\circ,60^\circ,114^\circ)$-similar triangles. (Reference: M. Laczkovich's Tilings of triangles.)

My attempt has been the following:

Since $\sin(42^\circ)$ and $\sin(78^\circ)$ have nested radicals, I tried to get rid of them by restricting my basic tiling units to only the $60^\circ$-angled isosceles trapezoids and parallelograms that are a single row of the triangular tiles. They have shorter-base-to-leg ratios of the form
$$m\cdot\frac{9-3\sqrt{5}}{2}+n\cdot\frac{11-3\sqrt{5}}{2}\quad\left(m,n\ge 0\right)$$
which are automatically algebraic integers. Any potential tiling of the equilateral triangle from these quadrilateral units corresponds to some integer polynomial relation of the above algebraics, whose polynomial degree correlates with the number of quadrilateral pieces in the tiling.

Unfortunately all the above algebraics have large norms, so a blind search for the desired polynomial is out of the question, and I had to reduce the pieces' proportions again to the rationals. I was able to find a $60^\circ$-angled isosceles trapezoid with shorter-base-to-leg ratio of $10$ using $79$ tiles, and a $60^\circ$-angled parallelogram with neighboring sides' ratio of $11$ using $80$ tiles. Thus a few more tiles produce a $60^\circ$-angled rhombus, and another few more tiles produce a $60^\circ$-angled isosceles trapezoid with shorter-base-to-leg ratio $1$, three of which tile an equilateral triangle, using a total of $121\,170$ triangular tiles. While I was at it, I found this less related post that might reduce my number of tiles to a bit below a hundred thousand.

Meanwhile, I also did a quick computer search through some conceptually simple configurations that attempt to tile the equilateral triangle using less than about $50$ tiles, and I found nothing at all.

I get this feeling that about a hundred thousand tiles is not the optimal amount for such a tiling, so I'm asking to see if people have better ideas. I'm unable to provide cash incentive as the parallel post did, but anyone who tries this puzzle will sure have my gratitude.

Edit suggested by RavenclawPrefect:

To get to the quadrilateral tiling units that I used, the first thing is to de-nest the radicals as I mentioned above. As $\mathbf{Q}(\zeta_{60})$ is Galois over $\mathbf{Q}(\sqrt{3})$ (the base field here should not be $\mathbf{Q}$ but instead the coordinate field of the equilateral triangle), if we can geometrically construct any length $\ell$ (or techinically, ratio $\ell$), such that when we perform the same geometric construction but with all the $42^\circ$ angles and $78^\circ$ angles swapped with each other, we still provably construct the same $\ell$, then it must hold that $\ell\in\mathbf{Q}(\sqrt{5})$, so that $\ell$ doesn't contain any nested radicals.

There were a couple of ideas on what $\ell$ should specifically be, most of them parallel ideas that can all be found in the parallel question for the square. I settled on the above $\mathbf{Q}(\sqrt{5})$-quadrilaterals (the ones that are a single row of triangular tiles) because they had the smallest numerator norms among others. As a non-example, there was a double-decker idea using $9$ tiles that resulted in a trapezoid with ratio a rational multiple of $889-321\sqrt{5}$, yuck. There were also some non-triviality in which way the triangles should be oriented when being put into a single row, but some more calculation showed that the above $(m,n)$ form are all we really get. More precisely, a trapezoid also can't have $m=0$, and a parallelogram also can't have $n=0$.

After all that work, the rest has really been a matter of trial-and-error. Among all the $(m,n)$ form, I picked a parallelogram with the smallest norm, which is an $(m,n)=(0,1)$ parallelogram with $4$ tiles, and rotated it so that it becomes a $\frac{11+3\sqrt{5}}{38}$-parallelogram. Then $19$ of those make a $\frac{11+3\sqrt{5}}{2}$-parallelogram with $76$ tiles, and obviously I combined it to a $(1,0)$-trapezoid and a $(0,1)$-parallelogram to get to the rational quadrilaterals.

So the process was more like "I frankly don't know what else to do" rather than "I see potential simplifications but I don't know the optimum". It's also why I'm seeking for completely new ideas (see above) that aren't found in the parallel question about the square.

RavenclawPrefect also asked a well-motivated question for if the same tiling could be performed but with congruent tiles. M. Laczkovich proved this is impossible in a subsequent paper Tilings of Convex Polygons with Congruent Triangles.

答:
I'm posting a new answer to this question, because the techniques I'm using differ substantially from the previous answer and it was already getting quite long. (Much of this answer was written prior to Anders' excellent answer, so it retreads some ground there.)

Edit: This answer in turn has been updated many, many times with improved constructions - feel free to peruse the edit history to see my earlier attempts.

To start with, I'd like to better flesh out the constructions outlined in the OP, as I found looking at these diagrams helpful. Define a parallelogram of ratio $r$ as one with sides $1,r,1,r$ in cyclic order, and a trapezoid of ratio $r$ as one with sides $1,r,1,r+1$ in cyclic order. (I'll implicitly assume that everything has $60^\circ$ and $120^\circ$ angles and that all trapezoids are isosceles unless otherwise stated.)

Here is an isosceles trapezoid of ratio $\frac{9-3\sqrt{5}}2$ made from three $\color{blue}{42}-\color{green}{60}-\color{red}{78}$ triangles:

Here is a parallelogram of a ratio $1$ larger (so with the same base) made from four such triangles:

(Note that it is not given by adding a triangle to the previous construction! The bottom three points are in the same location, though.)

As Edward H observes, we can actually extend either of the two parallelograms above by inserting a non-$60$-degree parallelogram in between an edge where only red and blue angles meet; this lets us spend $2$ more triangles to create trapezoids and parallelograms of ratios $\frac{9-3\sqrt{5}}{2}$ more.

Now, some observations:

A parallelogram of ratio $r$ is also a parallelogram of ratio $1/r$: just turn it on its side!

Given two parallelograms of ratios $r,s$, we can put them together to get a parallelogram of ratio $r+s$.

Given a trapezoid of ratio $r$ and a parallelogram of ratio $s$, we can put them together to get a trapezoid of ratio $r+s$.

Given two trapezoids of ratios $r,s$, we can turn one of them upside down and then put them together to get a parallelogram of ratio $r+s+1$ (because the top side is one unit shorter than the bottom side).

Given two trapezoids of ratios $r,s$, we can put one on top of another to get a trapezoid of ratio $rs/(r+s+1)$.

This gives us an obvious path forward: start with our two basic trapezoid and parallelogram solutions (plus their extensions), then combine them in the above ways looking for small tilings of nice rational-ratio trapezoids and parallelograms until we find a set we can nicely fill an equilateral triangle with.

I wrote up some code to perform exact computations with elements of $\mathbb{Q}[\sqrt{5}]$, and started storing all of the trapezoids and parallelograms one can make with up to around $90$ triangles, but bounding the size of the rational numbers involved to prevent the search space from getting too out of hand. (If I have a parallelogram of ratio $1173/292-46\sqrt{5}/377$, I'm probably not going to end up needing it.)

This alone doesn't turn up very many rational-ratio shapes, so I ran a second script that checked amongst all of the shapes generated in the previous iteration for those whose irrational parts were the negatives of each other, and combined them into new, rational-ratio shapes.

The results of this search included many interesting constructions, including Anders Kaseorg's 72-triangle solution for a parallelogram of unit ratio, but for our purposes we can focus on three of them: an $89$-tile trapezoid of ratio $3$, a $97$-tile trapezoid of ratio $2$, and a $113$-tile trapezoid of ratio $3/2$, shown below from top to bottom:

If we place the ratio-2 trapezoid on top of the ratio-3 trapezoid, we obtain a trapezoid of ratio $1$ using $186$ triangles:

Triplicating this would let us obtain an equilateral triangle with $558$ triangles, but we can do better using the ratio-$3/2$ trapezoid. Observe that three trapezoids form an equilateral triangle if and only if their ratios multiply to $1$. So, by stacking a ratio-$2$ trapezoid on a ratio-$3$ trapezoid, we obtain a ratio $2/3$ trapezoid, which combined with the ratio-$3/2$ trapezoid and the ratio-$1$ trapezoid forms an equilateral triangle.

Here is an image of the full (asymmetric!) construction, with all $(89+97)+(97+113)+113=\textbf{509}$ triangles in one piece:


答:
Hint:

Among the numbers $3,3^2,.., 3^{2012}$ there exists two with the same remainder when divided by $2011$.

Or

Since $2011$ is prime, by Fermat Little Theorem

$$3^{2010} \equiv 1 \pmod {2011}$$

Thus $3^{2010+k}-3^k$ is divisible by $2011$.

Title: Difference of two powers of $3$ divisible by $2011$

Body: How to prove that there exists two powers of $3$ that differ by a number that is divisible by $2011$?</p>


问:
Question about $\partial_i \left(f \circ x^{-1}\right)(x(p))$
In this video(at about 55:00), the lecturer wrote $\partial_i \left(f \circ x^{-1}\right)(x(p))$ as $\left (\dfrac{\partial f}{\partial x^{i}} \right)_p$. I am very confused about this notation. The lecturer says its not a partial derivative, but how is it useful to just denote like this? $x$ is a function in $\partial_i \left(f \circ x^{-1}\right)(x(p))$ but how can it be like a coordinate in $\left (\dfrac{\partial f}{\partial x^{i}} \right)_p$? Does it have something to do with partial derivatives? I would appreciate it if there's someone could clarify this for me.

答:
It is useful at least in two aspects.

When the smooth manifold is $\mathbb{R}^n$ and if one considers the identity map as the coordinate map, then $\partial_i(f\circ x^{-1})(x(p))$ coincide with the partial derivative in the classical sense.

For general manifolds, this notation reminds you of the name of the coordinate map and the direction of the partial derivative for $(f\circ x^{-1})$.

By definition of the video,
$$
\left(\frac{\partial f}{\partial x^i}\right)_p:=\partial_i(f\circ x^{-1})(x(p))
$$

What the lecturer means is that it is not a partial derivative for the function $f$, because it does not make sense to talk about "partial derivatives" for functions defined on a general manifold $M$; for instance, you cannot "add" two elements on $M$. But it is a partial derivative for the function $f\circ x^{-1}$, which is a function on the Euclidean space.


问:
Calculate:$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$
How to calculate following with out using L'Hospital rule

$$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$$

答:
\begin{align}
L &=\lim_{x \rightarrow (-1)^{+}}\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}\\
&=\lim_{t \rightarrow 0^{+}}\frac{\sqrt{\pi}-\sqrt{\cos^{-1}(t-1)}}{\sqrt{t}}\\
&= \lim_{t \rightarrow 0^{+}}\frac{\sqrt{\pi}-\sqrt{\pi}+\frac{\sqrt{t}}{\sqrt{2\pi}}+o(t)}{\sqrt{t}}\\
&= \frac{1}{\sqrt{2\pi}}
\end{align}


问:
How is y(x) and y similar?
In my differential equations textbook, there is a solution of a differential equation as $y(x)=2x+3+c$. From different values of $c$, a family of solutions is obtained. What confuses me is the graph on these calculus books where they have  plotted these solutions on $xy$-plane supposing $y$ as a variable or output of function which is actually a name of function. Shouldn't they have done $u=y(x)$ and then only plot $u$ vs $x$ for the graph of a function obtained?

答:
It is a matter of syntax:

For instance, we can write $y = x^2 + 5x-2$ or $y(x) = x^2 + 5x-2$.

The syntax may be differing; however, the semantics are the same. Using $y(x)$ stresses the fact that $y$ is dependent upon $x$ whereas the other does so implicitly . In my experience, $y$ is the convention outside of differential equations but have seen the notation $y(x)$ when in the context of differential equations.  It is not dissimilar to the usage of $y$ vs $f(x)$.

As mentioned in the comments $y$ and $y(x)$ can be used interchangeably.

Title: How is y(x) and y similar?


问:
How to Solve Large Scale Linear Least Squares with Simplex Constraints
I need to solve a rather simple optimization problem, however the structure of the problem makes out-of-the-box commercial solvers fail. The problem is:\begin{aligned}
&\min_{x_i}\|\sum_{i=1}^{10}x_i\mathbf{a}_i - \mathbf{b}\|_2^2 \\
&s.t.: \sum_{i=1}^{10}x_i = 1,\; x_i\geq0
\end{aligned}
The decision variables are $x_i$, which are scalars, and typically the problem will have 10 variables. The complexity comes from the fact that $\mathbf{a}_i, \mathbf{b}$ are VERY LONG constant vectors, varying from 10,000 to 100,000 elements and possibly more. Without going too much into details, black-box solvers fail since they impute the variable over vectors, thus converting a 10-D problem to 100K-vars problem.

Since the problem has a simple structure (only complicated by the constant vectors), I think that interior point algorithms can perform well here. Also, the Hessian is very simple and easy to compute - it's $ij$-th entry is simply $\langle \mathbf{a}_i, \mathbf{a}_j \rangle$, with size $10\times 10$. I'm very familiar with optimization, however my experience is mostly with first-order algorithms. I'm almost clueless when it come to IPMs and not sure how to implement an appropriate algorithm. Does anyhow have a ready script I can use? Or any other tips that might help?

答:
The dimensions of $A$ and $b$ are not relevant if you do some preprocessing:

$$||Ax-b||^2 = x^T(A^TA)x - 2 b^TAx+b^Tb$$

You can compute $A^TA$ and $A^Tb$ which are $10 \times 10$ and $10\times 1$ and specify the objective function with the formula on the right hand side, so the optimization algorithm never sees the 100k data points.


问:
Number of binary strings without adjacent char more than $n$
$N$ is a natural number, let $0\leq\alpha\leq 1$ a fixed real number, I want to estimate the number of length-N binary strings which statisfy: the same character (i.e $0$ or $1$) could not occur repeatedly and subsequently more than $\alpha N$ times.

For example: substrings like $000....000$ (more than $\alpha N$ many $0$) could not exist.

I only need to know how the number $I(N)$ of these strings grows when $N\to \infty$.

答:
The answer is close to $2^n$. Here's an easy computation which suggests something like this: for a random binary string, the expected number of times $k$ consecutive zeroes or ones appear is $\frac{2(n-k+1)}{2^k}$. Already from this simple calculation we expect that a "typical" binary string has runs of length about $\log_2 n$. It turns out to be unlikely for a binary string to have runs much longer than this, which can be proven as follows.

The number $I(n, k)$ of length-$n$ binary strings not containing $k$ consecutive $0$s or $1$s can be computed as follows. We can encode such a string by recording first its initial digit and second a word of length $n-1$ using letters $D$ and $S$ (for "different" and "same") which record whether each successive digit is the same or different than the previous one. This is readily seen to be a bijection, and moreover the original string does not contain $k$ consecutive $0$s or $1$s iff the new string does not contain $k-1$ consecutive $S$'es. The first digit is irrelevant. It follows that

$$I(n, k) = 2 J(n-1, k-1)$$

where $J(n-1, k-1)$ is the number of length-$(n-1)$ strings of $D$s and $S$'es not containing $k-1$ consecutive $S$'es. The behavior of this sequence is thoroughly analyzed in Example V.4 of Flajolet and Sedgewick's Analytic Combinatorics on page 308, which contains in particular the following estimate: if $L$ denotes the random variable given by the longest run of $S$'es in a random string of length length $n$, then

$$\mathbb{P}(L < \lfloor \log_2 n \rfloor + h) = \exp(-f(n) 2^{-h-1}) + O \left( \frac{\log n}{\sqrt{n}} \right)$$

where $f(n) = 2^{ \{ \log_2 n \} }$ and $\{ x \}$ denotes the fractional part; in particular $1 \le f(n) \le 2$. Replacing $n$ with $n-1$ and setting $h = k - \lfloor \log_2 n \rfloor$ gives

$$\frac{I(n, k)}{2^n} = \frac{J(n-1, k-1)}{2^{n-1}} = \exp \left( -f(n-1) 2^{-k - \lfloor \log_2 n \rfloor - 1} \right) + O \left( \frac{\log n}{\sqrt{n}} \right)$$

from which we conclude that for $\alpha \in (0, 1]$, $I(n, \alpha n)$ is close to $2^n$ as $n \to \infty$; their ratio is asymptotically $1 + O \left( \frac{\log n}{\sqrt{n}} \right)$ (for $k$ this large the exponential term is $1$ plus a term so small that the error term dominates), which in particular approaches $1$.

The weaker conclusion that $\lim_{n \to \infty} \frac{I(n, \alpha n)}{2^n} = 1$ can probably be proven by estimating the variance of the number of times $k$ consecutive $S$'es appear and then applying Chebyshev's inequality. Flajolet and Sedgewick give an asymptotic expansion for the expected length $\mathbb{E}(L)$ of the longest run (which is $\log_2 n + O(1)$ but they give a more precise asymptotic than this) and mention that the variance is $O(1)$.


答: If you divide by 2011 then there are 2011 possible remainders, if you look at the sequence 3, 3^2, ... 3^2012 then there are 2012 members in this sequence so by the pigeonhole princple two of them must have the same remainder on division by 2011, and hence there difference must be divisible by 2011.</p>


问:
How to show that a Brownian bridge is a Markov process?
If $W(t)$ is a standard Wiener process (i.e., for $t \geq 0, W(t)$ is normally distributed with expected value 0 and variance $t,$ and the increments are stationary and independent), then
$$
B(t)=W(t)-\frac{t}{T} W(T)
$$
is a Brownian bridge for $t \in[0, \mathrm{T}]$.

How to show that $B$ is a Markov process ? I have 2 ways in mind: some sort of direct proof or showing that $B$ is an Ito process (then it is automatically Markov). I would like to see both.

答:
Edit: this is not correct by @UBM's comment

We have $dB_t = dW_t - \frac{W(T)}{T}dt$, so $B_t$ is an Ito process.

To prove directly that $B_t$ is Markov is essentially the same. Note that $B_t$ is a sum of a Wiener process and a term that depends exclusively on time. The latter term is clearly independent of the filtration, since it is deterministic. The former is known to be independent of the filtration since it has independent increments. It isn't too difficult to convince oneself that in this case this makes $B_t$ independent of the filtration.


问:
Are convexity and concavity sufficient or necessary condition for $\min \max = \max \min$
If $f$ is convex in $x$ and concave in $y$, then $\min_{x} \max_{y} f(x,y) = \max_{y} \min_{x} f(x,y)$. Is that a sufficient or necessary condition?
My intuition tells that it is a necessary but from my understanding from reading literature is sufficient since the equality (i.e., strong duality) can also happen if Slater's condition is met. If is it necessary, is there a connection between Slater's condition and a saddle point?

答:
The condition is neither necessary nor sufficient.

For equality to hold, just having a convex-concave function is not enough. You need something more, e.g., that the domain of either $X$ or $Y$ is compact, or that the equality comes from an optimization problem for which Slater's condition holds. A simple counterexample when no regularity condition is met, is $f(x,y) = xy$, for which $\min_x \max_y f(x,y) = \infty$ while $\max_y \min_x f(x,y) = -\infty$.

On the other hand, you can have equality even when $f$ is not convex-concave. A trivial example is a constant function defined on just two points, e.g., $f : \{0,1\} \to \mathbb{R}$ with $f(0)=f(1)=0$.


问:
How does subtracting an exponent from an exponent return a greater value?
I'm just stuck on a problem where I have to simplify an expression. This is the expression:

$\sqrt{x^2\:-\:\left(\frac{x}{2}\right)^2}$

The textbook has the answer as:

$\frac{\sqrt{3}x}{2}$

I have no idea how to get to that. I've tried to figure it out for the past hour and I have tried online solvers but I can't understand what the steps are. Any help would be very much appreciated.

Thanks!

答:
One step a time!

You solution assumes that $x\ge 0$.

\begin{align}
x^2-(\frac{x}{2})^2
&=x^2-\frac{x^2}{4}\\
&=1\cdot x^2-\frac{1}{4}\cdot x^2\\
&=(1-\frac{1}{4})x^2=
\frac34 x^2
\end{align}

So
$$
\sqrt{x^2-(\frac{x}{2})^2}=\sqrt{\frac34 x^2}
=\frac{\sqrt{3}}{\sqrt{4}}\sqrt{x^2}=\frac{\sqrt{3}}{2}x\quad x\ge 0
$$


问:
Why not define $i=\sqrt{-1}$ and use that to construct complex numbers?
I've seen the multitude of questions here about defining complex numbers, but I haven't found the answer I'm looking for there:

In elementary algebra, we typically get to $x^2=-1$, define $i=\sqrt{-1}$, and call it a day. In analysis, we define the complex numbers as ordered pairs, along with addition and (a strange-looking) multiplication, then derive the fact that $i^2=-1$. The only places I've seen attempt to justify this simply state it's "more mathematical" to do the latter.

My question is: why do we need to define the complex number system before defining $i$? Why not state $i \triangleq \sqrt{-1}$ and then construct the complex numbers starting from that definition?

答:
Strictly speaking, you can't just make definitions and hope they work. Practically speaking though, mathematicians have a lot of machinery built precisely to make sense of definitions that basically go, "Hm, I wish I had a element that had this property" - and these sorts of definition are ubiquitous in many branches of math. This extends well beyond defining arithmetic on the complex numbers, but let me briefly explain how one may define $\mathbb C$ using the machinery that exists for algebraic structures.

There's a pretty common way to define the complex numbers that basically just says "$\mathbb C$ can be constructed by adding to $\mathbb R$ an extra element $i$ such that $i^2=-1$." You usually avoid square roots in any definition since they're a bit ambiguous. This definition reads as follows:
$$\mathbb C = \mathbb R[i]/(i^2+1)$$
This is a ring theoretic construction, so let me unpack it: we're going to construct $\mathbb C$ from $\mathbb R$ in two steps:

Consider first the polynomial ring $\mathbb R[i]$. This consists of every polynomial in some variable $i$ - so has terms such as $i^3+i^2+17i+289$ or $i^{100}$ or just real constants such as $\pi$ - all with the usual rules for adding and multiplying. Polynomial rings capture the idea of "adding" or "adjoining" a new variable to a given ring without specifying any particular properties thereof. If you wanted to add a new variable, but didn't know anything about it, this is what you'd get.

The $/(i^2+1)$ part of the definition says: Take that polynomial ring, and now consider two polynomials $P(i)$ and $Q(i)$ to be equivalent if their difference $P(i)-Q(i)$ is a multiple of $i^2+1$. This is a bit of a tricky thing to get your head around, but note that $i^2$ is equivalent to $-1$ since their difference is $1\cdot (i^2+1)$. Similarly, $i^4$ is equivalent to $1$ since the difference satisfies the relation:
$$i^4-1=(i^2-1)(i^2+1)$$
where these are equations of polynomials (and are true of any variable or value as input for $i$).

The second rule, as it turns out, makes every polynomial equivalent to one of the form $a+bi$ - if you have a quadratic or higher term, you can find some multiple of $i^2+1$ with the same leading term as your quadratic, and subtract this multiple from your original polynomial, leading to an equivalent polynomial of lesser degree. Algorithmically, if you wanted to figure out what $i^3+2i^2$ is, you would first subtract $i\cdot (i^2+1)=i^3+i$ from this to get $2i^2-i$, then subtract $2(i^2+1)$ from that to get that your original polynomial $i^3+2i^2$ is equivalent to $-i-2$. This process is known as polynomial long division.

Note that addition and multiplication rules can then be derived - for instance
$$(a+bi)+(c+di)=(a+b)+(c+d)i$$
is just ordinary addition of polynomials. For multiplication you get:
$$(a+bi)(c+di)=ac+(ad+bc)i+bdi^2$$
as polynomials and then you can subtract $bd(i^2+1)$ (which is equivalent to $0$) to get:
$$(a+bi)(c+di)=(ac-bd)+(ad+bc)i$$
in $\mathbb C$.

It turns out that this definition is really the same (in a sense that can be made precise as a universal property in category theory) as saying (where a "ring" is just a structure in which addition and multiplication are sensibly defined):

$\mathbb C$ is a ring containing $\mathbb R$ and an extra element $i$ where the only equalities between elements are those that can be proven from the assumption that $i^2+1=0$.

And the prior equations can be interpreted in this context too - if we knew that
$i^2=-1$, deducing that $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ would simply be done by distribution followed by a substitution. The explicit ring theoretic construction is mostly needed to ensure that there actually is some sensible $\mathbb C$ that realizes this property - however, these sorts of constructions exist in great generality; if all you want to do is add, subtract, and multiply (but not necessarily divide), you can basically add however many variables you like and demand that they satisfy as many polynomial equations as you want, and you'll get, well, something out - though that thing can be badly behaved in many ways (even in ways as extreme as "oops, everything equals zero now"). The complex numbers, however, turn out to be a very well-behaved example of such a construction, fundamentally because $i^2+1$ cannot be factored as a real polynomial.

It's a bit more tricky when you want to talk about analysis and topology (since this construction only deals with algebraic not analytical properties), but this is not a particularly great obstacle - this definition can be refined in various ways to preserve those kinds of structures too.


答:
Here is an answer which addresses the underlying logical issues that your question raises.

I could define a jackalope to be be a rabbit with antelope horns, but then the trouble arises whether any rabbit with antelope horns exists, because if not then my definition is vacuous nonsense.

Similarly, you could define $i$ to be $\sqrt{-1}$, but then you would have to contend with the problem of whether $\sqrt{-1}$ exists. In the real number system it is a theorem that $\sqrt{-1}$ does not exist, or to be more precise there is no real number solution to the equation $x^2=-1$.

So, until someone produces a reasonable number system in which $x^2=-1$ has a solution, the definition $i=\sqrt{-1}$ is vacuous nonsense.

Fortunately, someone has done exactly that, namely produced a number system (in fact a field) in which $x^2=-1$ exists. Namely, those ordered pairs of real numbers with that strange looking multiplication formula that you mentioned in your post.

Title: Why not define $i=\sqrt{-1}$ and use that to construct complex numbers?

Body:
I've seen the multitude of questions here about defining complex numbers, but I haven't found the answer I'm looking for there:

In elementary algebra, we typically get to $x^2=-1$, define $i=\sqrt{-1}$, and call it a day. In analysis, we define the complex numbers as ordered pairs, along with addition and (a strange-looking) multiplication, then derive the fact that $i^2=-1$. The only places I've seen attempt to justify this simply state it's "more mathematical" to do the latter.

My question is: why do we need to define the complex number system before defining $i$? Why not state $i \triangleq \sqrt{-1}$ and then construct the complex numbers starting from that definition?


问:
Defining ordered pairs using only first-order logic
The usual definition of an ordered pair $\langle x,y \rangle$ is $\{\{x\},\{x,y\}\}$, but how may one define that using purely first-order logic contructs (variables, logical operations, quantifiers) and the $\in$ relation?

After reading the wiki, which defines the properties:

$x$ is the first coordinate of $p$: $\forall A[A \in p \rightarrow x \in A]$

$x$ is the second coordinate of $p$: $\exists A[A \in p \land x \in A] \land \forall A \forall B[A \in p \land B \in p \rightarrow (A \neq B \rightarrow x \notin A \lor x \notin B)]$

I arrived at the following definition:

$$
\langle x,y \rangle = p \leftrightarrow \forall A[A \in p \rightarrow x \in A] \land \exists A[A \in p \land y \in A] \land \forall A \forall B[A \in p \land B \in p \rightarrow (A \neq B \rightarrow y \notin A \lor y \notin B)]
$$

However, this definition would also consider $\{\{x\},\{x,y\},\{x,z\}\}$ or $\{\{x\},\{x,y,z\}\}$ as representing that ordered pair, and as such, it is not unique.

So, how may one define ordered pairs using only first-order logic such that only the set $\{\{x\},\{x,y\}\}$ represents the ordered pair $\langle x,y \rangle$?

答:
$$
\newcommand\liff{\leftrightarrow}
\newcommand\lif{\rightarrow}
\newcommand\lfi{\leftarrow}
\newcommand\ordp[2]{\langle #1,#2 \rangle}
\newcommand\mset[1]{\{ #1 \}}
\newcommand\isRel[1]{#1 \text{ is a relation}}
\newcommand\isFunc[1]{#1 \text{ is a function}}
\newcommand\isOneOne[1]{#1 \text{ is one-one}}
\newcommand{\fitch}[1]{\begin{array}{rlr}#1\end{array}}
\newcommand{\fcol}[1]{\begin{array}{r}#1\end{array}} %FirstColumn
\newcommand{\scol}[1]{\begin{array}{l}#1\end{array}} %SecondColumn
\newcommand{\tcol}[1]{\begin{array}{l}#1\end{array}} %ThirdColumn
\newcommand{\subcol}[1]{\begin{array}{|l}#1\end{array}} %SubProofColumn
\newcommand{\startsub}{\\[-0.29em]} %adjusts line spacing slightly
\newcommand{\endsub}{\startsub} %adjusts line spacing slightly
\newcommand{\fendl}{\\[0.044em]} %adjusts line spacing slightly
$$

$$\ordp{a}{b} = p \iff \forall Z[Z \in p \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]]$$

In this definition, $p$ is our ordered pair, and every set $Z$ in $p$ has to be either $\mset{a}$ or $\mset{a,b}$. This is more explicit with the equivalent set of definitions:

$$
\mset{a} = p \iff \forall x[x \in p \liff x = a] \\
\mset{a,b} = p \iff \forall x[x \in p \liff x = a \lor x = b] \\
\ordp{a}{b} = p \iff \forall Z [Z \in p \liff Z=\mset{a} \lor Z=\mset{a,b}] \\
$$

The former is more in line with the question, since it is a single definition and uses only primitive symbols. However, I found the latter nicer to work with, as it makes proofs a lot less verbose. In any case, they are equivalent, so a proof that is valid for one is valid for the other.

We may prove that set-theory guarantees their unique existence. The first definition will be used in this example.

The following statements trivially follow from Paring, Specification and Extensionality:
$$
\begin{array}{l}
\text{[1]}:\forall a \exists A \forall x[x \in A \liff x = a] \\
\text{[2]}:\forall a \forall b \exists A \forall x[x \in A \liff x = a \lor x = b] \\
\text{[3]}:\forall p \forall q \forall a[\forall x[x \in p \liff x = a] \land \forall x[x \in q \liff x = a] \lif p = q] \\
\text{[4]}:\forall p \forall q \forall a \forall b[\forall x[x \in p \liff x = a \lor x = b] \land \forall x[x \in q \liff x = a \lor x = b] \lif p = q] \\
\end{array}
$$

We may use the first two to introduce the sets $A_{\alpha a}$ and $A_{\beta ab}$, which represent $\mset{a}$ and $\mset{a,b}$, and then introduce $p_{\alpha ab}$, which represents our ordered pair:

$$
\small
\fitch{
 \fcol{1:\fendl 2:\fendl 3:\fendl 4:\fendl 5:\fendl 6:\fendl 7:\fendl \vdots\fendl
 }
 & \scol {
  \exists A \forall x[x \in A \liff x = a] \\
  \exists A \forall x[x \in A \liff x = a \lor x = b] \\
  \forall x[x \in A_{\alpha a} \liff x = a] \\
  \forall x[x \in A_{\beta ab} \liff x = a \lor x = b] \\
  \exists A \forall x[x \in A \liff x = A_{\alpha a} \lor x = A_{\beta ab}] \\
  \forall x[x \in p_{\alpha ab} \liff x = A_{\alpha a} \lor x = A_{\beta ab}] \\
  Z \in p_{\alpha ab} \liff Z = A_{\alpha a} \lor Z = A_{\beta ab} \\
  \rlap{\vdots} \hphantom{\subcol{\subcol{\forall x[x \in Z \liff x=a \lor x=b] \land \forall x[x \in A_{\beta ab} \liff x=a \lor x=b] \lif Z = A_{\beta ab}}}} \\
 }
 & \tcol{
  [1]\ \forall\text{E} \fendl
  [2]\ \forall\text{E} \fendl
  1\ \exists\text{E} \fendl
  2\ \exists\text{E} \fendl
  [2]\ \forall\text{E}[a/A_{\alpha a},b/A_{\beta ab}] \fendl
  5\ \exists\text{E}[A/p_{\alpha ab}] \fendl
  6\ \forall\text{E}[x/Z] \fendl
  \rlap{\vdots} \hphantom{aaa\ \forall\text{E}[a/A_{\alpha a},b/A_{\beta ab}]} \fendl
 }
}
$$

Then, we prove that $p_{\alpha ab}$ follows the defined properties, which means that a being with these properties exists (line $36$):

$$
\small
\fitch{
 \fcol{\vdots\fendl 8:\fendl 9:\fendl 10:\fendl 11:\fendl 12:\fendl 13:\fendl 14:\fendl 15:\fendl 16:\fendl 17:\fendl 18:\fendl 19:\fendl 20:\fendl 21:\fendl 22:\fendl 23:\fendl 24:\fendl 25:\fendl 26:\fendl 27:\fendl 28:\fendl 29:\fendl 30:\fendl 31:\fendl 32:\fendl 33:\fendl 34:\fendl 35:\fendl 36:\fendl \vdots\fendl
 }
 & \scol {
  \rlap{\vdots}
  \startsub\subcol{
   Z \in p_{\alpha ab} \\
   \hline
   Z = A_{\alpha a} \lor Z = A_{\beta ab}
   \startsub\subcol{
    Z = A_{\alpha a} \\
    \hline
    \forall x[x \in Z \liff x = a] \\
    \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b] \\
   }\endsub
   Z = A_{\alpha a} \lif \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]
   \startsub\subcol{
    Z = A_{\beta ab} \\
    \hline
    \forall x[x \in Z \liff x = a \lor x = b] \\
    \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b] \\
   }\endsub
   Z = A_{\beta ab} \lif \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b] \\
   \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b] \\
  }\endsub
  Z \in p_{\alpha ab} \lif \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]
  \startsub\subcol{
   \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]
   \startsub\hline\subcol{
    \forall x[x \in Z \liff x = a] \\
    \hline
    \forall x[x \in Z \liff x=a] \land \forall x[x \in A_{\alpha a} \liff x=a] \lif Z = A_{\alpha a} \\
    Z = A_{\alpha a} \\
    Z = A_{\alpha a} \lor Z = A_{\beta ab} \\
   }\endsub
   \forall x[x \in Z \liff x = a] \lif Z = A_{\alpha a} \lor Z = A_{\beta ab}
   \startsub\subcol{
    \forall x[x \in Z \liff x = a \lor x = b] \\
    \hline
    \forall x[x \in Z \liff x=a \lor x=b] \land \forall x[x \in A_{\beta ab} \liff x=a \lor x=b] \lif Z = A_{\beta ab} \\
    Z = A_{\beta ab} \\
    Z = A_{\alpha a} \lor Z = A_{\beta ab} \\
   }\endsub
   \forall x[x \in Z \liff x = a \lor x = b] \lif Z = A_{\alpha a} \lor Z = A_{\beta ab} \\
   Z = A_{\alpha a} \lor Z = A_{\beta ab} \\
   Z \in p_{\alpha ab}
  }\endsub
  \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b] \lif Z \in p_{\alpha ab} \\
  Z \in p_{\alpha ab} \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b] \\
  \forall Z[Z \in p_{\alpha ab} \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]] \\
  \exists p \forall Z[Z \in p \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]] \\
  \rlap{\vdots} \\
 }
 & \tcol{
  \rlap{\vdots} \hphantom{aaa\ \forall\text{E}[a/A_{\alpha a},b/A_{\beta ab}]} \fendl
  \text{P}[Z] \fendl
  7,8\ {\liff}\text{E} \fendl
  \text{P}[Z] \fendl
  3,10\ {=}\text{S}[A_{\alpha a}/Z] \fendl
  11\ {\lor}\text{I} \fendl
  10,12\ {\lif}\text{I}\ \fendl
  \text{P}[Z] \fendl
  4,14\ {=}\text{S}[A_{\beta ab}/Z] \fendl
  15\ {\lor}\text{I} \fendl
  14,16\ {\lif}\text{I} \fendl
  9,13,17\ {\lor}\text{E} \fendl
  8,18\ {\lif}\text{I} \fendl
  \text{P}[Z,a,b] \fendl
  \text{P}[Z,a,b] \fendl
  [3]\ \forall\text{E}[p/Z,q/A_{\alpha a}] \fendl
  3,21,22\ {\lif}\text{E} \fendl
  22\ {\lor}\text{I} \fendl
  20,23\ {\lif}\text{I} \fendl
  \text{P}[Z,a,b] \fendl
  [4]\ \forall\text{E}[p/Z,q/A_{\beta ab}] \fendl
  4,26,27\ {\lif}\text{E} \fendl
  27\ {\lor}\text{I} \fendl
  25,28\ {\lif}\text{I} \fendl
  20,25,30\ {\lor}\text{E} \fendl
  7,31\ {\liff}\text{E} \fendl
  20,32\ {\lif}\text{I} \fendl
  19,33\ {\liff}\text{I} \fendl
  34\ \forall\text{I}\ \fendl
  35\ \exists\text{I}\ \fendl
  \rlap{\vdots} \hphantom{aaa\ \forall\text{E}[a/A_{\alpha a},b/A_{\beta ab}]} \fendl
 }
}$$

Then we prove that this being is unique, by showing that if any two sets $p$ and $q$ follow these properties, they are one and the same (line $45$):

$$
\small
\fitch{
 \fcol{\vdots\fendl 37:\fendl 38:\fendl 39:\fendl 40:\fendl 41:\fendl 42:\fendl 43:\fendl 44:\fendl\fendl\fendl 45:\fendl\fendl\fendl 46:\fendl
 }
 & \scol {
  \rlap{\vdots} \hphantom{\subcol{\subcol{\forall x[x \in Z \liff x=a \lor x=b] \land \forall x[x \in A_{\beta ab} \liff x=a \lor x=b] \lif Z = A_{\beta ab}}}}
  \startsub\subcol{
   \forall Z[Z \in p \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]] \\
   \forall Z[Z \in q \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]] \\
   \hline
   Z \in p \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b] \\
   Z \in q \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b] \\
   Z \in p \liff Z \in q \\
   \forall Z[Z \in p \liff Z \in q] \\
   p = q \\
  } \endsub
  \forall Z[Z \in p \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]] \\\quad\ \land \forall Z[Z \in q \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]] \\\quad \lif p = q \\
  \forall p \forall q[\forall Z[Z \in p \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]] \\\quad\ \land \forall Z[Z \in q \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]] \\\quad \lif p = q] \\
  \exists p! \forall Z[Z \in p \liff \forall x[x \in Z \liff x = a] \lor \forall x[x \in Z \liff x = a \lor x = b]] \\
 }
 & \tcol{
  \rlap{\vdots} \hphantom{aaa\ \forall\text{E}[a/A_{\alpha a},b/A_{\beta ab}]} \\
  \text{P}[p,a,b] \fendl
  \text{P}[q,p,a,b] \fendl
  37\ \forall\text{E} \fendl
  38\ \forall\text{E} \fendl
  39,40\ {\liff}\text{Tr} \fendl
  41\ \forall\text{I} \fendl
  42,\text{Ext.}\ \forall\text{I} \fendl
  37,38,43\ {\lif}\text{I} \fendl\fendl\fendl
  45\ \forall\text{I} \fendl\fendl\fendl
  36,45\ \exists!\text{I} \fendl
 }
}
$$

As such, there exists a unique $p$ with these properties (line $46$).

We may also prove that they obey the characteristeric property for ordered pairs, namely: $\forall x \forall y \forall a \forall b[\ordp{x}{y} = \ordp{a}{b} \liff x = a \land y = b]$. The second definition will be used in this example.

The following statements trivially follow from the definitions of $\mset{a}$ and $\mset{a,b}$, and Extentionality:
$$
\begin{array}{l}
\text{[1]}:\forall a \forall b [\mset{a} = \mset{b} \liff a=b] \\
\text{[2]}:\forall a [\mset{a,a} = \mset{a}] \\
\text{[3]}:\forall a \forall b [\mset{a,b} = \mset{a} \liff a=b] \\
\text{[4]}:\forall a \forall b \forall c [a \neq b \lif \mset{c} \neq \mset{a,b}] \\
\text{[5]}:\forall a \forall b \forall c [\mset{a,b} = \mset{a,c} \liff b = c] \\
\end{array}
$$

We start by proving the trivial implication from RHS to LHS:

$$
\small
\fitch{
 \fcol{
  1:\fendl 2:\fendl 3:\fendl 4:\fendl \vdots\fendl
 }
 & \scol {
  \subcol{
   x=a \land y=b \\
   \hline
   \ordp{x}{y} = \ordp{x}{y} \\
   \ordp{x}{y} = \ordp{a}{b} \\
  } \endsub
  x=a \land y=b \lif \ordp{x}{y} = \ordp{a}{b} \\
  \rlap{\vdots} \hphantom{\subcol{\mset{x,y}=\mset{a} \lor \mset{x,y}=\mset{a,b} \liff \mset{x,y}=\mset{x} \lor \mset{x,y}=\mset{x,y}}} \\
 }
 & \tcol{
  \text{P}[x,y,a,b] \fendl
  \text{T} \fendl
  1,2\ {=}\text{S} \fendl
  1,3\ {\lif}\text{I} \fendl
  \rlap{\vdots}\hphantom{\text{Def.}\ \forall\text{E}[a/x,b/y,p/\ordp{a}{b}]} \fendl
 }
}
$$

For the trickier LHS to RHS, we start by assuming that $\ordp{x}{y} = \ordp{a}{b}$ and using the definition to establish a relationship between the sets (line $14$). From that relationship, we extract some useful statements about them (lines $19$ to $22$):

$$
\small
\fitch{
 \fcol{
  \vdots\fendl 5:\fendl 6:\fendl 7:\fendl 8:\fendl 9:\fendl 10:\fendl 11:\fendl 12:\fendl 13:\fendl 14:\fendl 15:\fendl 16:\fendl 17:\fendl 18:\fendl 19:\fendl 20:\fendl 21:\fendl 22:\fendl \vdots\fendl
 }
 & \scol {
  \subcol{
   \rlap{\vdots} \\
   \ordp{x}{y} = \ordp{a}{b} \\
   \hline
   \ordp{a}{b} = \ordp{x}{y} \liff \forall Z [Z \in \ordp{x}{y} \liff Z=\mset{a} \lor Z=\mset{a,b}] \\
   \ordp{x}{y} = \ordp{a}{b} \liff \forall Z [Z \in \ordp{a}{b} \liff Z=\mset{x} \lor Z=\mset{x,y}] \\
   \forall Z [Z \in \ordp{x}{y} \liff Z=\mset{x} \lor Z=\mset{x,y}] \\
   \forall Z [Z \in \ordp{a}{b} \liff Z=\mset{a} \lor Z=\mset{a,b}] \\
   Z \in \ordp{x}{y} \liff Z=\mset{a} \lor Z=\mset{a,b} \\
   Z \in \ordp{a}{b} \liff Z=\mset{x} \lor Z=\mset{x,y} \\
   Z \in \ordp{a}{b} \liff Z=\mset{a} \lor Z=\mset{a,b} \\
   Z=\mset{a} \lor Z=\mset{a,b} \liff Z=\mset{x} \lor Z=\mset{x,y} \\
   \forall Z [Z=\mset{a} \lor Z=\mset{a,b} \liff Z=\mset{x} \lor Z=\mset{x,y}] \\
   \mset{a}=\mset{a} \lor \mset{a}=\mset{a,b} \liff \mset{a}=\mset{x} \lor \mset{a}=\mset{x,y} \\
   \mset{x}=\mset{a} \lor \mset{x}=\mset{a,b} \liff \mset{x}=\mset{x} \lor \mset{x}=\mset{x,y} \\
   \mset{a,b}=\mset{a} \lor \mset{a,b}=\mset{a,b} \liff \mset{a,b}=\mset{x} \lor \mset{a,b}=\mset{x,y} \\
   \mset{x,y}=\mset{a} \lor \mset{x,y}=\mset{a,b} \liff \mset{x,y}=\mset{x} \lor \mset{x,y}=\mset{x,y} \\
   \mset{a}=\mset{x} \lor \mset{a}=\mset{x,y} \\
   \mset{x}=\mset{a} \lor \mset{x}=\mset{a,b} \\
   \mset{a,b}=\mset{x} \lor \mset{a,b}=\mset{x,y} \\
   \mset{x,y}=\mset{a} \lor \mset{x,y}=\mset{a,b} \\
   \rlap{\vdots}
  }
 }
 & \tcol{
  \rlap{\vdots}\hphantom{\text{Def.}\ \forall\text{E}[a/x,b/y,p/\ordp{a}{b}]} \fendl
  \text{P}[x,y,a,b] \fendl
  \text{Def.}\ \forall\text{E}[p/\ordp{x}{y}] \fendl
  \text{Def.}\ \forall\text{E}[a/x,b/y,p/\ordp{a}{b}] \fendl
  5,6\ {\liff}\text{E} \fendl
  5,7\ {\liff}\text{E} \fendl
  8\ \forall\text{E} \fendl
  9\ \forall\text{E} \fendl
  5,10\ {=}\text{S} \fendl
  11,12\ {\liff}\text{Tr} \fendl
  13\ \forall\text{I} \fendl
  14\ \forall\text{E}[Z/\mset{a}] \fendl
  14\ \forall\text{E}[Z/\mset{x}] \fendl
  14\ \forall\text{E}[Z/\mset{a,b}] \fendl
  14\ \forall\text{E}[Z/\mset{x,y}] \fendl
  15\ {\liff}\text{E} \fendl
  16\ {\liff}\text{E} \fendl
  17\ {\liff}\text{E} \fendl
  18\ {\liff}\text{E} \fendl
  \rlap{\vdots}\hphantom{\text{Def.}\ \forall\text{E}[a/x,b/y,p/\ordp{a}{b}]} \fendl
 }
}
$$

Now, either $a = b$ or $a \neq b$. So we may prove that in both cases it follows that $a=x$ and $b=y$. In the case that $a = b$:

$$
\small
\fitch{
 \fcol{
  \vdots\fendl 23:\fendl 24:\fendl 25:\fendl 26:\fendl 27:\fendl 28:\fendl 29:\fendl 30:\fendl 31:\fendl 32:\fendl 33:\fendl 34:\fendl 35:\fendl 36:\fendl 37:\fendl 38:\fendl 39:\fendl 40:\fendl 41:\fendl \vdots\fendl
 }
 & \scol {
  \subcol{
   \rlap{\vdots} \hphantom{\mset{x,y}=\mset{a} \lor \mset{x,y}=\mset{a,b} \liff \mset{x,y}=\mset{x} \lor \mset{x,y}=\mset{x,y}} \\
   a=b \lor a \neq b
   \startsub\subcol{
    a = b \\
    \hline
    \mset{x}=\mset{a} \lor \mset{x}=\mset{a,b} \\
    \mset{x}=\mset{a} \lor \mset{x}=\mset{a,a} \\
    \mset{a,a} = \mset{a} \\
    \mset{x}=\mset{a} \lor \mset{x}=\mset{a} \\
    \mset{x}=\mset{a} \\
    \mset{x}=\mset{a} \liff x=a \\
    x=a \\
    \mset{x,y}=\mset{a} \lor \mset{x,y}=\mset{a,b} \\
    x=b \\
    \mset{x,y}=\mset{x} \lor \mset{x,y}=\mset{x,x} \\
    \mset{x,x}=\mset{x} \\
    \mset{x,y}=\mset{x} \lor \mset{x,y}=\mset{x} \\
    \mset{x,y}=\mset{x} \\
    \mset{x,y}=\mset{x} \liff x=y \\
    x=y \\
    y=b \\
   } \endsub
   a=b \lif (x=a \land y=b) \\
   \rlap{\vdots} \hphantom{\mset{x,y}=\mset{a} \lor \mset{x,y}=\mset{a,b} \liff \mset{x,y}=\mset{x} \lor \mset{x,y}=\mset{x,y}} \\
  }
 }
 & \tcol{
  \rlap{\vdots}\hphantom{\text{Def.}\ \forall\text{E}[a/x,b/y,p/\ordp{a}{b}]} \fendl
  \text{T} \fendl
  \text{P}[x,y,a,b] \fendl
  20\ \text{C} \fendl
  24,25\ {=}\text{S}[b/a] \fendl
  [2]\ \forall\text{E} \fendl
  27,26\ {=}\text{S}[\mset{a,a}/\mset{a}] \fendl
  28\ {\lor}\text{E} \fendl
  [1]\ \forall\text{E}[a/x,b/a] \fendl
  29,30\ {\liff}\text{E} \fendl
  22\ \text{C}\ \fendl
  24,31\ {=}\text{Tr} \fendl
  32,33\ {=}\text{S}[b/x] \fendl
  [2]\ \forall\text{E}[a/x] \fendl
  34,35\ {=}\text{S}[\mset{x,x}/\mset{x}] \fendl
  36\ {\lor}\text{E} \fendl
  [3]\ \forall\text{E}[a/x,b/y] \fendl
  37,38\ {\liff}\text{E} \fendl
  33,39\ {=}\text{Tr} \fendl
  24,31,40\ {\lif}\text{I} \fendl
  \rlap{\vdots}\hphantom{\text{Def.}\ \forall\text{E}[a/x,b/y,p/\ordp{a}{b}]} \fendl
 }
}
$$

In the case that $a \neq b$:

$$
\small
\fitch{
 \fcol{\vdots\fendl 42:\fendl 43:\fendl 44:\fendl 45:\fendl 46:\fendl 47:\fendl 48:\fendl 49:\fendl 50:\fendl 51:\fendl 52:\fendl 53:\fendl 54:\fendl 55:\fendl 56:\fendl 57:\fendl 58:\fendl
 }
 & \scol {
  \subcol{
   \rlap{\vdots} \hphantom{\mset{x,y}=\mset{a} \lor \mset{x,y}=\mset{a,b} \liff \mset{x,y}=\mset{x} \lor \mset{x,y}=\mset{x,y}}
   \startsub\subcol{
    a \neq b \\
    \hline
    \mset{x}=\mset{a} \lor \mset{x}=\mset{a,b} \\
    a \neq b \lif \mset{x} \neq \mset{a,b}\\
    \mset{x} \neq \mset{a,b} \\
    \mset{x}=\mset{a} \\
    \mset{x}=\mset{a} \liff x=a \\
    x = a \\
    \mset{a,b}=\mset{x} \lor \mset{a,b}=\mset{x,y} \\
    \mset{a,b}=\mset{x,y} \\
    \mset{a,b}=\mset{a,y} \\
    \mset{a,b}=\mset{a,y} \liff b=y\\
    b = y
   } \endsub
   a \neq b \lif (x=a \land y=b)  \\
   x=a \land y=b
  } \endsub
  \ordp{x}{y} = \ordp{a}{b} \lif x=a \land y=b \\
  \ordp{x}{y} = \ordp{a}{b} \liff x=a \land y=b \\
  \forall x \forall y \forall a \forall b[\ordp{x}{y} = \ordp{a}{b} \liff x=a \land y=b] \\
 }
 & \tcol{
  \rlap{\vdots}\hphantom{\text{Def.}\ \forall\text{E}[a/x,b/y,p/\ordp{a}{b}]} \fendl
  \text{P}[x,y,a,b] \fendl
  20\ \text{C} \fendl
  [4]\ \forall\text{E}[c/x] \fendl
  42,44\ {\lif}\text{E} \fendl
  43,45\ {\lor}\text{E} \fendl
  [1]\ \forall\text{E}[a/x,b/a] \fendl
  46,47\ {\liff}\text{E} \fendl
  21\ \text{C} \fendl
  45,49\ {\lor}\text{E} \fendl
  48,50\ {=}\text{S}[x/a] \fendl
  [5]\ \forall\text{E}[c/y] \fendl
  51,52\ {\liff}\text{E} \fendl
  42,53\ {\lif}\text{I} \fendl
  23,41,54\ \text{TI} \fendl
  5,55\ {\lif}\text{I} \fendl
  4,56\ {\liff}\text{I}\ \fendl
  57\ \forall\text{I}\ \fendl
 }
}
$$

As such, the definitions obey the characteristic property for ordered pairs.


问:
Is the following version of the fundamental lemma of the calculus of variations valid?
Let $U$ be an open subset of $\mathbb{R}^n$ with smooth boundary $\partial U$. Consider a function $f$ in $L^2(u)$. Suppose that for every $h$ in the Sobolev space$ H^2_0(U)$ it holds that
$$\int_U f \Delta h=0.$$
Where $\Delta$ is the Laplacian operator.n

Can we conclude that $f=0$ almost everywhere on $U$?

答:
Let $u: \overline {U}\to \mathbb{R}$ be a harmonic function, i.e. $$\tag{1}\Delta u=0\ in\ U$$

Mutliply $(1)$ by $h\in H_0^2(U)$ in both sides and then integrate: $$\tag{2}\int_U\Delta u\cdot h=0,\ \forall\ h\in H_0^2(U)$$

Use the generalized Green identity to conclude from $(2)$ that $$\tag{3}0=\int_U\Delta u\cdot h=-\int_U\nabla u\nabla h=\int_Uu\Delta h,\ \forall\ h\in H_0^2(U)$$

From $(3)$ we have your claim but $u$ does not need to be zero almost everywhere.


问:
Derivative and partial derivative of complex functions
I know the formal definition of a derivative of a complex valued function, and how to compute it (same as how I would for real-valued functions), but after doing some problems, I feel as if I could just take the partial derivative w.r.t $x$ of the function to compute the derivative (so it doesn't depend on $y$?) as opposed to taking derivative w.r.t $z$ first then substitute.
That might be a bit obscure, so I'll put in a couple of examples

Examples

All of the examples are analytic (satisfy the Riemann conditions) with $z = x+iy$ and $f(z) = u(x,y)+iv(x,y)$.

 1. $f(z) = z = x+iy =u(x,y) + iv(x,y)$.

The derivative is $f'(z) = 1$. Another way would be just to take partials of $f(z)$ w.r.t $x$ to get the result.

$f(z) = z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$

$f'(z) = 2z = 2x + 2iy$

Another way is to just directly take partial derivative of $f$ w.r.t $x$ since $\frac{\partial u}{\partial x} = 2x$ and $\frac{\partial v}{\partial x} = 2y$.

$f(z) = z^3 = (x+iy)^3 = x^3 - 3xy^2 + i(3x^2 y - y^3)$.

$ f'(z) = 3z^2 = 3x^2 -3y^2 + 6ixy$.

This can also be found similarly in other examples since $\frac{\partial u}{\partial x} = 3x^2 - 6xy$ and $\frac{\partial v}{\partial x} = 6xy$.

So it seems that I could just take the partial derivatives with respect to $x$ of the resultant complex number, and ignore the $y$ to find the derivatives. How come this is true?

答:
There is a difference between a general function of two variables f(x,y) and a complex function f(z)=f(x+iy). In the first, the function variations with x and y are completely independent, whereas in a complex function the variable (x+iy) transforms wholly. Accordintly, whatever happens to x, happens to y too (and hence z as well) in terms of the slope, which is the (partial) derivative. Hope this helps.

Masoud


问:
Derivative of $-\csc(2x)$
Is the following reasoning correct?
$$-\csc2x = -1[\csc(2x)\cdot\cot(2x)] = -1[\csc(2x)\cdot\cot(2x)\cdot2\cdot2]$$

I am unsure whether that is correct so far because I do not know if the derivatives of the $\csc$ and $\cot$ need to be taken again due to the chain rule

答:
Why don't you simply use sine and cosine?

The derivative can then be easily calculated using basic differentiation rules:
$$
-\frac{d}{dx}\csc(2x)=\frac{d}{dx}\frac{-1}{\sin(2x)}=\frac{1}{\sin^2(2x)}\cdot \cos(2x)\cdot 2=2\cos(2x)\csc(2x)
$$

"Is the following reasoning correct?" NO. You miss the derivative symbol, not to say other aspects.

Title: Derivative of $-\csc(2x)$


问:
How can I derive posterior distribution of theta when having 2 R.V (X, y) using Bayesian Rule?
In the book, Machine Learning in Finance: From Theory to Practice(Matthew F. Dixon, et al, 2020), there is a bayesian equation derivation in Chapter2:

$p(\theta|y, X) = {p(\theta, y | X) \over p(y|X)} = {p(\theta)p(y|\theta, X) \over p(y|X)}$

I have two questions about this:

How can I interpret $p(\theta|y, X)$? Should I see it as $p(\theta|(y \cap X))$ or $p((\theta|y), X)$?

How is the above equation derived? I have no idea how $p(\theta|y, X)$ goes to ${p(\theta, y | X) \over p(y|X)}$ (why is it normalized(?) by X on both nominator and denominator? and why $p(y|X)$ should be denominator?, etc), and how ${p(\theta, y | X) \over p(y|X)}$ is converted to

${p(\theta)p(y|\theta, X) \over p(y|X)}$ suddenly(How can the $p(\theta)$ came out of $p(\theta, y | X)$?, etc)

答:
$P(\theta \mid y, X)$ denotes the conditional probability/density of $\theta$ given $y$ and $X$. $p((\theta \mid y), X)$ is definitely wrong. $p(\theta \mid (y \cap X))$ is in the right spirit, but I would not use the $\cap$ notation because $y$ and $X$ are presumably random variables, not events.

You should be familiar with $$p(\theta \mid y) = \frac{p(\theta, y)}{p(y)} = \frac{p(\theta) p(y \mid \theta)}{p(y)}.$$ This is simply the definition of conditional probability applied twice (and is often called Bayes's rule). What you have is a similar application, where each of the five terms appearing in this equation is also conditioned on $X$, i.e. $$p(\theta \mid y, X) = \frac{p(\theta, y \mid X)}{p(y \mid X)} = \frac{p(\theta \mid X) p(y \mid \theta, X)}{p(y \mid X)}.$$ If this is unclear to you, I would encourage you to just break every conditional probability down into unconditional probabilities, and check that the equalities work out i.e. $p(\theta \mid y, X) = \frac{p(\theta, y, X)}{p(y, X)}$, $p(y \mid X) = \frac{p(y, X)}{p(X)}$, and so on.

Finally, the reason why $p(\theta \mid X)$ is replaced by $p(\theta)$ is probably due to some context that you did not provide in the question. I presume that $y$ depends on both $\theta$ and $X$, but $\theta$ does not depend on $X$.


问:
Evaluating double integral by change in variable.
Evaluate the following integral by changing to polar coordinates:
$\iint x dxdy$, where $0\leq x \leq y$ and $0 \leq y \leq 1$.
Above integral can be evaluated directly without changing the variables. I am getting the answer $\frac{1}{6}$. But I have to evaluate it by changing in polar coordinates. So let $x=r\cos\theta$ and $y=r\sin\theta$. Then $dxdy=rdrd\theta$. But what will be the limits on $r$ and $\theta$ ?

答:
Hint: Draw the region is key to understand the integral. If you do the graph you will find $$\frac{\pi}{4}\leq \theta\leq \frac{\pi}{2}$$ and $$0\leq y\leq 1$$
$$0\leq r \text{sin}(x)\leq 1$$
$$0\leq r\leq\frac{1}{\text{sin}(x)}$$


答:
As I see the problem is that, after you extend the square root function from $[0,\infty )$ to $\{-1\}\cup [0,\infty )$ adding a symbol $i:=\sqrt{-1}$ to $\mathbb{R}$, you still need to construct a minimal field (or an algebra) that contains $\mathbb{R}\cup \{i\}$, and you still would need to define what is the value of $i^2$.

One approach could be to set a multiplication operation in the new space such that

$$
\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}\quad \text{ for all }\,a,b\in \mathbb{R}
$$

and this would imply that $i^2=1$, what would give an algebra on $\mathbb{F}:=\{a+bi: a,b\in \mathbb{R}\}$ (the Clifford algebra $\operatorname{Cl}_{0,1}(\mathbb{R})$) but not a field. Thus, by itself, setting the value of $i$ doesn't lead immediately to a value for $i^{2}$.


问:
Question about transcendental equation
I need to show that $f(x) = 1-e^{-2xf(x)}$ for $\lim_{x \to 1/2} f(x) = 0$ and $\lim_{x \to \infty} f(x) = 1$ is monotonically increasing on the intervall $[1/2, \infty[$. I can't seem to extract any information off of the limits. I know that $$f'(x) = \frac{2f(x)e^{-2xf(x)}}{1-2xe^{-2xf(x)}}.$$ But looking at curvature around 1/2 doesn't help aswell since the limit for $f''$ can't be calculated with what is given. Am I missing something?

答:
You xan work this problem making $f(x)$ explicit.
$$f(x) = 1-e^{-2xf(x)}\implies f(x)=1+\frac 1{2x}W\left(-2x\, e^{-2 x}\right)$$ where appears Lambert function.

For more legibility, I shall use $t=-2x\, e^{-2 x}$

From this
$$f(x)=1+\frac{W(t)}{2x}$$
$$f'(x)=W(t)\frac{ x t'-t (1+W(t))}{2 x^2 t(1+W(t))}$$
$$f''(t)=W(t)\frac {-x^2 W(t) (W(t)+2) t'^2+x t (W(t)+1)^2 \left(x t''-2
   t'\right)+2 t^2 (W(t)+1)^3 }{2 x^3 t^2 (1+W(t))^3 }$$

Title: Question about transcendental equation


问:
Difference of two powers of $3$ divisible by $2011$
How to prove that there exists two powers of $3$ that differ by a number that is divisible by $2011$?</p>

答: Hint: $2$ of the $2012$ integers $\ 3^0, 3^1, \ldots, 3^{2011}\ $ have the same remainder when divided by $2011,\,$ i.e. the pigeonholes are the $2011$ possible remainders modulo $2011$.</p>


问:
How to cut a cube out of a tree stump, such that a pair of opposing vertices are in the center?
I saw this picture of a cube cut out of a tree stump.

I've been trying to craft the same thing out of a tree stump, but I found it hard to figure out how to do it.

One of the opposing vertices pair is on the center of the tree stump:

I've been struggling to find the numbers and angle needed to make the cuts.

Any kind of advice would be greatly appreciated. Thanks for taking your time to read this and I'm sorry for my bad english.

答:
The Method

Here's an approach with which you are only ever dealing with planar surfaces.

Cut from the stump a (regular) hexagonal prism of radius $r$ and height $\frac32r\sqrt2\approx2.1213 r$.

Note: The face-to-opposite-face "width" of the prism should be $r\sqrt3\approx1.732r$.

(Sanity check for regularity: The edges of the hexagonal top and bottom should also be $r$.)

The figure shows the top view and the six "unrolled" lateral faces.

Divide each lateral edge of the prism into thirds, and draw a zig-zag across the lateral faces, joining alternating $\frac13$ and $\frac23$ marks on those edges.

Each segment of the zig-zag will be an edge of the final cube. (Its length should be $e\approx 1.2247r$.)

Each consecutive triad of points on the zig-zag (along with either the top-center or bottom-center point of the prism) defines a face of the cube, so "all you have to do" is remove excess wood until those faces are planar.

Done!

The Math

For a prism with radius $r=2$ (to avoid fractions), let the bottom-center of the prism be the origin $O=(0,0,0\sqrt2)$, and the top-center be $T=(0,0,3\sqrt2)$.

The coordinates of the zig-zag of points ($ABCDEFA$ in the figure) have the form
$$\left(2\cos k\,60^\circ, 2\sin k\,60^\circ, (1\;\text{or}\;2)\sqrt{2} \right)$$
Specifically,
$$A = (\phantom{-}2,0,2\sqrt{2}) \qquad B = (\phantom{-}1,\phantom{-}\sqrt{3},1\sqrt2) \qquad C = (-1,\phantom{-}\sqrt3,2\sqrt2)$$
$$D = (-2,0,1\sqrt2) \qquad E = (-1,-\sqrt3,2\sqrt2) \qquad F = (\phantom{-}1,-\sqrt3,1\sqrt2)$$
(where $A$, $C$, $E$ are closer to the top of the prism, and $B$, $D$, $F$ are closer to the bottom). From here, we can verify that
$$\begin{align}
T &= B+(A-B)+(C-B) &&\quad\to\quad T,A,B,C\;\text{coplanar} \\
&= D+(C-D)+(E-D)   &&\quad\to\quad T,C,D,E\;\text{coplanar}\\
&= F+(E-F)+(A-F)   &&\quad\to\quad T,E,F,A\;\text{coplanar}\\[4pt]
O &= A+(B-A)+(F-A) &&\quad\to\quad O,F,A,B\;\text{coplanar}\\
&=C+(B-C)+(D-C)    &&\quad\to\quad O,B,C,D\;\text{coplanar}\\
&=E+(D-E)+(F-E)    &&\quad\to\quad O,D,E,F\;\text{coplanar}
\end{align}$$
and
$$\begin{align}2\,\sqrt{\frac32}=\sqrt{6}\;&=\;|AB|=|BC|=|CD|=|DE|=|EF|=|FA| \\
&=\;|TA|=|TC|=|TE| \\
&=\;|OB|=|OD|=|OE|\end{align}$$
so that the faces are all quadrilaterals with equal edge-lengths, making them at least rhombuses. Also,
$$\begin{align}
0\;&=\;(A-T)\cdot(C-T)=(C-T)\cdot(E-T)=(E-T)\cdot(A-T) \\
&=\;(B-O)\cdot(D-O)=(D-O)\cdot(F-O)=(F-O)\cdot(B-O)
\end{align}$$
so that
$$90^\circ = \angle ATC = \angle CTE = \angle ETA =\angle BOD = \angle DOF = \angle FOB$$
That is, each rhombus face has at least one right angle, making it a square. $\square$


问:
Compute the $n$-th power of the difference of the following matrices.
Given the matrices $$A=\begin{pmatrix} a & c & b \\ c & b & a \\ b & a & c \end{pmatrix}$$ and $$B=\begin{pmatrix} b & a & c \\ a & c & b \\ c & b & a \end{pmatrix}$$

for $a,b,c \in \mathbb{R}$
find $$(A-B)^n$$ for $n \in \mathbb{N}$.

I can see that the matrices are symmetric and also they have the same determinant since $B$ can be obtained by swapping the rows of $A$ twice.

I attempted to solve this by using the Cayley-Hamilton theorem.
First I find the characteristic polynomial which is $$p(k)=-k^3+(a+b+c)k^2-(ab+bc+ac)k+3abc-b^3-a^3-c^3$$
which is the same for both matrices.

Now by Cayley-Hamilton theorem, I have $$-A^3+(a+b+c)A^2-(ab+bc+ac)A+3abc-b^3-a^3-c^3=0$$
and $$-B^3+(a+b+c)B^2-(ab+bc+ac)B+3abc-b^3-a^3-c^3=0$$
By subtracting those two equations by parts, I get $$-A^3+B^3+(a+b+c)(A^2-B^2)-(ab+bc+ac)(A-B)+3abc-b^3-a^3-c^3$$

and this is where I am stuck, I attempted to use the identities $$a^3+b^3+c^3-3abc=\frac{1}{2} (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$
and $$2ab+2bc+2ca=(a+b+c)^2-a^2-b^2-c^2$$ but couldn't end up with something useful.

My question is, first of all, is my approach correct? How should I proceed?

答:
Note that $M = A - B$ can be written in the form
$$
\pmatrix{
p&r&q\\
r&q&p\\
q&p&r},
$$
where $p+q+r = 0$. Verify that
$$
M \pmatrix{1\\1\\1} = 0,
$$
which means that $\det(A - B)$ must be zero. With that, we find the characteristic polynomial to be
$$
\det(M - kI) = -k^3 - (pq + pr + qr)k.
$$
Let $\alpha = pq + pr + qr$. The Cayley Hamilton theorem tells us that
$$
-M^3 - \alpha M = 0 \implies M^3 = -\alpha M.
$$
With that established, we can conclude that for $n \geq 2$ we have
$$
M^n = \begin{cases}
\alpha^{n/2 - 1} M^2 & n \text{ is even,}\\
\alpha^{(n-1)/2} M & n \text{ is odd.}
\end{cases}
$$