问:
How to prove $\left \lceil \frac{n}{m} \right \rceil = \left \lfloor \frac{n+m-1}{m} \right \rfloor$?
everybody, how can I prove that, for natural $m$ and $n$,
$$
\left \lceil \frac{n}{m} \right \rceil = \left \lfloor \frac{n+m-1}{m} \right \rfloor \; ?
$$
Thanks a lot.</p>

答:
This is basically Bill's solution, I remembered a pretty nice way of writing it:

Let

$f : {\mathbb Z}_+ \rightarrow {\mathbb R}$ defined by

$$f(x)= \left \lceil \frac{x}{m} \right \rceil - \left \lfloor \frac{x+m-1}{m} \right \rfloor \,.$$

Then

$$f(x+m)= \left \lceil \frac{x+m}{m} \right \rceil - \left \lfloor \frac{x+2m-1}{m} \right\rfloor =\left \lceil \frac{x}{m} \right \rceil +1- \left \lfloor \frac{x+m-1}{m}\right\rfloor -1=  f(x) $$

Thus, $f$ is periodic of period $m$.

Moreover $f(0)=0$ and for all $1 \leq x \leq m-1 \,;\, f(x)=1-1=0$.

Since $f$ is periodic and identically zero on a period, $f$ is the zero function.

Last but not least, it is easy to see that if we replace $x \in {\mathbb Z}$ by $x$ positive real, the only step which fails is that $f(x) \neq 0$ for $x \in (0,1)$.

So unless I made a mistake the equality holds on $[0, \infty) \backslash [(0,1)+ {\mathbb Z}]$.


问:
Equation of Partial derivatives
I am currently learning multivariable calculus and the following was done in my lecture to prove one of the properties of Jacobian

The property:

($JJ'=1$), where $J$ is $\frac{\partial (x,y)}{\partial(u,v)}$ and $J'$ is $\frac{\partial(u,v)}{\partial (x,y)}$

The equation in the proof:

$\partial u = \frac{\partial u}{\partial x} \partial x +\frac{\partial u}{\partial y} {\partial y}$

Then the professoor proceeded to divide by $\partial u$ on both sides so you end up with $\frac{\partial u}{\partial u}$ on the left hand side which is essential 1. He then did this for $v$ also.

My question: $\partial u\ , \partial x\ , \partial y$ on their own makes no sense. And since partial derivative is defined for suppose u with respect to some other variable, dividing or multiplying by $\partial (some \ variable)$ should be wrong. What is the correct theory ?

答:
You are correct: the notation $\partial u$, etc, is not really clear. Dividing the equation in the question by $\partial u$ would give, formally,
$$
\frac{\partial u}{\partial u}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial u}
$$
which doesn't make much sense.

Here is how I would handle this:

The Chain Rule says
$$
\left.\begin{align}
\frac{\mathrm{d}u}{\mathrm{d}t}&=\frac{\partial u}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial u}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}\\[3pt]
\frac{\mathrm{d}v}{\mathrm{d}t}&=\frac{\partial v}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial v}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}
\end{align}\right\}\hspace{.5cm}
\begin{bmatrix}\dfrac{\mathrm{d}u}{\mathrm{d}t}\\\dfrac{\mathrm{d}v}{\mathrm{d}t}\end{bmatrix}=\begin{bmatrix}
\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\
\dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y}
\end{bmatrix}
\begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}\tag1
$$
and
$$
\left.\begin{align}
\frac{\mathrm{d}x}{\mathrm{d}t}&=\frac{\partial x}{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}+\frac{\partial x}{\partial v}\frac{\mathrm{d}v}{\mathrm{d}t}\\[3pt]
\frac{\mathrm{d}y}{\mathrm{d}t}&=\frac{\partial y}{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}+\frac{\partial y}{\partial v}\frac{\mathrm{d}v}{\mathrm{d}t}
\end{align}\right\}\hspace{.5cm}
\begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}=\begin{bmatrix}
\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\
\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}
\end{bmatrix}
\begin{bmatrix}\dfrac{\mathrm{d}u}{\mathrm{d}t}\\\dfrac{\mathrm{d}v}{\mathrm{d}t}\end{bmatrix}\tag2
$$
Plugging $(1)$ into $(2)$ gives
$$
\begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}
=\begin{bmatrix}
\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\
\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}
\end{bmatrix}
\begin{bmatrix}
\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\
\dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y}
\end{bmatrix}
\begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}\tag3
$$
That is,
$$
\begin{bmatrix}
\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\
\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}
\end{bmatrix}
\begin{bmatrix}
\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\
\dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y}
\end{bmatrix}
=I\tag4
$$

ContentLicense: CC BY-SA 4.0

Title: Equation of Partial derivatives


问:
Is there a reasonably accurate and easy way to approximate lbs-stones or kg-stones (and vice versa)?
For example, when I need to talk with my American friends about body weight:

$\mathrm{KG}\times2+10\%$

($100\times2=200$, $200+10\%=220$... Pretty accurate and easy to do mentally, both ways)

However, when I need to talk with my British friends about body weight, how do I approximate from KG to Stones?

Additionally, if I have a similar formula for Stones-lbs, that will be useful too!

I mean, I know a Stone is 14 lbs, but how to approximate 247lbs in Stones without pulling out a calculator?

For your reference, a Stone is about 6.35 KG

I am thinking that a "good, practical" approximation follows these rules:

5% accuracy is kinda bad... Imagine weighing 100kg and wanting to lose 5kg - the error alone is making discussion about it difficult! I am thinking up to 2% error

Maximum of 3 steps

Only addition, subtraction, multiplication and division allowed (% also allowed)

Multiplication only by "easy" factors: 2, 3, 4, 5, 10

Division only by "easy divisors": 2, 10

Only "easy" percentages like: 1%, 10%...

答:
20 stones are equal to 127 kilograms.

This is the best I could get and it's not good enough.

For the reverse (kgs to stones) there's no easy way, it seems.

You cannot rely there would be an easy conversion simply because

the metric system is not compatible with the British one

(in terms of "good" rational conversions).

So you cannot find easy to remember whole numbers $A$ and $B$ such that

$A$ stones are equal to $B$ kilograms. Let alone doing it both ways.

So in general you do need a calculator or... just wait for the British/US to adopt widely the metric system


问:
Is this power series expansion of $2{({\sqrt{\frac{1}{x}}})^M}{K_M}({2\sqrt{\frac{1}{x}}})$ correct?
I am calculating the power series approximation of $f\left( x \right)$ at $x = \infty$.

With my
$f\left( x \right) = 2{\left( {\sqrt {\frac{1}{x}} } \right)^M}{K_M}\left( {2\sqrt {\frac{1}{x}} } \right)$

Where ${K_M}\left( x \right)$ is the modified Bessel function of the second kind of order $M$ (cf. http://dlmf.nist.gov/10.32.E9). In my application, M is a positive integer.

However, the received answer seem to be strange and I do not know if it correct or not ?

Because for any positive integer $M$ then $\Gamma \left( { - M} \right)$ would result in a Complex Infinity.

However, when I plug in the actual value $M=2,4$. I receive a neat approximation that suit my application. Therefore, I hope to be able to derive a generalized answer like this for any $M$ (even or odd).

Please help me with this !

Thank you !

答:
In the output line, it starts with $x^{\color{red}{-}M}$ so, who cares since $M>0$ and $x\to \infty$.

So, just use
$$\Gamma (M)\left(1-\frac{1}{(M-1) x}+\frac{1}{2 (M-2) (M-1)
   x^2}+O\left(\frac{1}{x^3}\right)\right)$$

Edit

The case $M=1$ is totally different. The result should be
$$1-\frac{\log (x)-2 \gamma +1}{x}-\frac{2 \log (x)-4 \gamma +5}{4 x^2}+O\left(\frac{1}{x^3}\right)$$


问:
Is there a continuous extention at $(0,0)$ $f(x,y)= \frac{x^{2021} *e^y}{1+x-e^y}$
$$f(x,y)= \frac{x^{2021} *e^y}{1+x-e^y}$$

i said that domain is {$(x,y) \in R^2 : x+e^y \neq -1$}

I tried to show that yes then no but i did not succeed

答: Hint: $|f(\frac 1  n , \ln (1+\frac1  n+\frac 1 {n^{2022}})| \to \infty$ as $ n \to \infty$.


问:
How to represent $ \frac{1}{a_n}$ in terms of the generating function of $a_n$?
Suppose I have the ordinary generating function of $a_n$,

$$ A(x) = \sum_{i} a_i x^i$$

Then from the above expression, is there a way to write the generating function of $\frac{1}{a_n}$?

If there is a way to do it using exponential generating function or dirchlet generating function, then that too is fine.

答:
My guess is that is not possible to get a closed form.

Take $a_n = (n+1)^2$. Then $\displaystyle A(x) = \sum_{n=0}^\infty a_n x^n =\frac{1+x}{(1-x)^3}$ and $\displaystyle B(x) = \sum_{n=0}^\infty\frac{1}{a_n}x^n = \frac{1}{x}\rm{Li}_2(x)$ where $\rm{Li}_2$ is the polylogarithm of order $2$. If there were a closed form for $B(x)$ in terms of $A(x)$ one would have a closed form for $\rm{Li}_2$, but there is no known closed form for $\rm{Li}_2$.

And this is just an example taking a very simple sequence $a_n$. If less trivial sequences are considered many others series whose values are unknown will be automatically found, which is extremely unlikely.


问:
Prove certain set is convex
I'm having trouble proving this statement:

Prove that the set
$$X=\{ (x,y)\in\mathbb{R}^2 : y\geq 0, x\leq b+y \}$$
given $b\in\mathbb{R}$ is convex.

The work I've done so far:

I first assumed we havo two different points $(x_1,y_1),(x_2,y_2)\in X$. Then, the segment between them is made of the points
$$(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2) \ , \ \lambda\in[0,1].$$
So if I prove that points are inside $X$ for any $\lambda$ I would have proven the convexity.
It's easy to see that

$$\lambda y_1 + (1-\lambda)y_2 \geq 0,$$
since it's a positive number plus another positive number. Now, I don't know how to prove that

$$\lambda x_1 + (1-\lambda)x_2 \leq b + \lambda y_1 + (1-\lambda)y_2$$
knowing obviously that $x_1 \leq b+y_1$ and $x_2\leq b+y_2$.
What can I do to prove this part? Any help or hint will be appreciated, thanks in advance.

答: Just multiply $x_1 \leq b+y_1$ by $\lambda$, $x_2 \leq b+y_2$ by $(1-\lambda)$ and add the inequalities. Finally note that $\lambda b +(1-\lambda) b=b$.

Title: Prove certain set is convex


问:
How many arrangements? $(a, b, c, d, x, y, z, w)$
I would appreciate if somebody could help me with the following problem:

Q: How many arrangements?

My Study

I solved by mathematica11.0

```<code>f[n_] := {a, b, c, d, x, y, z, w} /.
  Solve[0 < a < b < c < d < n && 0 < x < y < z < w < n && a < x &&
    b < y && c < z && d < w, {a, b, c, d, x, y, z, w}, Integers]
f[9]=490
</code>```

答:
This is only a partial solution.

The problem is somehow related to Young tableaux.  If we were required to use all the numbers from $1$ to $8$, we could compute the number of tableau of this shape with the hook-length formula which in this instance gives $$\frac{8!}{5\cdot4\cdot3\cdot2\cdot4\cdot3\cdot2\cdot1}=14$$

Here are the hook lengths:
$$5\ 4\ 3\ 2\\
  4\ 3\ 2\ 1$$
It is apparent that we can have any or all of the equalities $b=x,\ c=y, d=z$. For example, if
$$abcd\\
xyzw$$  is a standard tableau, we could replace it by either $$abcd\\byzw$$ or $$axcd\\xyzw$$  For each of the $3$ diagonals, we have $3$ choices: leave it unchanged, or make it constant in one of two ways, which brings us up to $27\cdot14=378$ solutions.

So far all our solutions have $a=1,\ w=8$, but there are others, for example $$2345\\3456$$  I haven't been able to think of a simple way to count these.

I've Googled "Young tableaux with repeated elements," but I didn't find anything bearing directly on this problem.  Admittedly, I've just spent a few minutes on this.  This paper discusses a number of variants, but a brief perusal didn't turn up anything directly relevant.


答:
Let (x₁,y₁),(x₂,y₂) ∈ X
Let (c,d) = (λx₁ + (1-λ)x₂, λy₁ + (1-λ)y₂)
c = λx₁ + (1-λ)x₂ ≤ λ(b + y₁) + (1-λ)(b + y₂)
= (λb + b - λb) + λy₁ + (1-λ)y₂
= b + d
hence c ≤ b + d
and you have already proved why d ≥ 0

It's actually pretty straight forward. You should've just expanded the expression.

Title: Prove certain set is convex

Body:
I'm having trouble proving this statement:

Prove that the set
$$X=\{ (x,y)\in\mathbb{R}^2 : y\geq 0, x\leq b+y \}$$
given $b\in\mathbb{R}$ is convex.

The work I've done so far:

I first assumed we havo two different points $(x_1,y_1),(x_2,y_2)\in X$. Then, the segment between them is made of the points
$$(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2) \ , \ \lambda\in[0,1].$$
So if I prove that points are inside $X$ for any $\lambda$ I would have proven the convexity.
It's easy to see that

$$\lambda y_1 + (1-\lambda)y_2 \geq 0,$$
since it's a positive number plus another positive number. Now, I don't know how to prove that

$$\lambda x_1 + (1-\lambda)x_2 \leq b + \lambda y_1 + (1-\lambda)y_2$$
knowing obviously that $x_1 \leq b+y_1$ and $x_2\leq b+y_2$.
What can I do to prove this part? Any help or hint will be appreciated, thanks in advance.


问:
Find the missing angle in triangle
In the below triangle, we are looking for the value of angle $φ$.

We are given $α=30, β=18, γ=24$ and also that $CD=BD$.

I have solved it with trigonometry (sine law) and found the required angle to be 78 but I need to solve it with Geometry only.

What I have tried so far:

First of all, the angle is constructible, which means to me that there must be a geometrical solution. I first drew triangle ABC; easy, since we know 2 of its angles. We are not interested in the lengths of the sides. Then, with side AC as a base, and angle of 24 degrees, we can draw a ray from the point A.

Then, since $CD=BD$, triangle DCB is isosceles, therefore D must lie on the perpendicular bisector of CB, which we can draw. The point of intersection of the ray from A and the perpendicular bisector, is point D.

From triangle FEB we have that

angle AFD = 108.

From triangle AFD,

$ADC+CDE+54+108=180$ so $ADC+CDE=18$

We also have $24+ACD+ADC=180$

$ACB=132$

$132+φ+ACD=180$

$18+φ+54+ADC+2CDE=180$

I am always one equation short.

Any ideas?

Many thanks in anticipation!

EDIT:

Sine law in triangle ABD:

$\frac {sin (φ+18)}{AD} = \frac {sin (54)}{BD}$

Sine law in triangle ACD:

$\frac {sin (360-132-φ)}{AD} = \frac {sin (24)}{CD} = \frac {sin (24)}{BD}$

so

$\frac {sin (φ+18)}{sin (228-φ)} = \frac {sin (54)}{sin (24)}$

hence $φ=78$.

答:
Consider a regular $30$-gon $X_1X_2X_3X_4X_5X_6X_7X_8X_9X_{10}X_{11}X_{12}X_{13}X_{14}X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}X_{22}X_{23}X_{24}X_{25}X_{26}X_{27}X_{28}X_{29}X_{30}$ and place it on the plane so that $X_1 \equiv A$, $X_6\equiv B$, and that $X_2$ and $C$ lie on different halfplanes determined by the line $AB$. Denote $K=X_2$, $L=X_3$, $M=X_4$, $N=X_5$, and $X_{15}=R$.

Build regular pentagon $KLOPQ$ as in the picture. We shall prove that $P\equiv C$.

Note that $\angle QKA = \angle LKA - \angle LKQ = 168^\circ - 108^\circ = 60^\circ$. Since $QK=KL=AK$, it follows that the triangle $AKQ$ is equilateral. In particular, $AQ=KQ=QP$, so $Q$ is the circumcenter of $AKP$. Angle chasing yields $\angle AQP = 360^\circ - 2\angle PKA = 360^\circ - 2(60^\circ + 36^\circ) = 168^\circ$, so by SAS triangle $AQP$ is congruent to $KLM$, $MNB$, and by symmetry it is congruent to $MOP$. Continuing angle chasing, $\angle PAQ = 6^\circ$, and finally $\angle BAP = \angle KAQ - \angle PAQ - \angle KAB = 60^\circ - 6^\circ - 24^\circ = 30^\circ$.

On the other hand, by congruency of $KLM$, $MNB$ and $MOP$, we have $MK=MP=MB$, so $M$ is the circumcenter of $KPB$ and therefore $\angle BMP = 2\angle BKP = 2(\angle LKP - \angle LKB) = 2(72^\circ - 18^\circ) = 108^\circ$, hence $\angle PBM = 36^\circ$ and $\angle PBA = \angle PBM - \angle ABM = 36^\circ - 18^\circ = 18^\circ$.

Since $\angle BAP = 30^\circ$ and $\angle PBA = 18^\circ$, we have that $P\equiv C$.

We shall prove now that $R\equiv D$. First of all, we have $\angle CAR = \angle BAR - \angle BAC = 54^\circ - 30^\circ = 24^\circ$. Secondly, since $\angle LKC = 72^\circ = \angle LKR$, we have that $K$, $C$, $R$ are collinear. Since $M$ is the circumcenter of $CKB$, we have $\angle BCR = \frac 12 \angle BMK = \frac 12 \cdot 156^\circ = 78^\circ$. We also have $\angle RBC = \angle RBA - \angle CBA = 96^\circ - 18^\circ = 78^\circ$. Since $\angle BCR = \angle RBC$, it follows that $R$ lies on the perpendicular bisector of $CB$, which along with $\angle CAR = 24^\circ$ means that $R\equiv D$. The answer follows:
$$\varphi = \angle BCD = \angle BCR = 78^\circ.$$

Title: Find the missing angle in triangle


问:
Why use Transpose notation in standart basis vectors unit length equation?
I had thought I understood everything about orthonormality until I encountered the notation on the source I share below. Why add another step using T? In other sources, it worked just fine without. Could you explain the reason why a human being would spend extra bits and energy to add that?

答:
For two column vectors $\vec x$ and $\vec y$ of the same dimension, we have $$\vec x\cdot \vec y=\vec x^T\vec y$$
based on the definition of matrix multiplication and the dot product. The transpose notation is more preferable for someone who exclusively wants to use matrix operations instead of vector operations like dot products.


问:
Are there any examples of non-computable real numbers?
Is this true, that if we can describe any (real) number somehow, then it is computable?

For example, $\pi$ is computable although it is irrational, i.e. endless decimal fraction. It was just a luck, that there are some simple periodic formulas to calcualte $\pi$. If it wasn't than we were unable to calculate $\pi$ ans it was non-computable.

If so, that we can't provide any examples of non-computable numbers? Is that right?

The only thing that we can say is that these numbers are exist in many, but we can't point to any of them. Right?

答: Examples of uncomputable numbers, like the one clearly presented by Luis Fonsera, seem to assume that the halting of any particular Turing machine can be determined (by some other Turing machine ?). Is this always true? It not then for some values of n, Fonsera's function f(n) might not be determined and so the uncomputable number r is undetermined. An example might be a Turing machine that searches for violations of the Goldbach conjecture. Another could be a Turing machine, T, that searches through ratios of two  rational numbers to see if it can find one that is squareroot of 2. We know that this programme wil never halt but I am at a loss to see how a Turing machine could come to show this just by examining the programme of machine T. Are some mathematical truths like the irrationality of squareroot 2  uncomputable?</p>


问:
What should I do when the factor $a_n$ in a power series is a series itself?
We were given the following exercise:

calculate the convergence radius of the following series and examine their behaviour on the edge of their respective convergence circles:
$$\sum_{n=0}^{\infty}(1+\frac 1{8}+\frac 1{27}+\ldots+\frac 1{n^{3}})x^n$$
$$\sum \frac {x^{n^2}}{2^n}$$

now correct me if I'm wrong, but you can calculate a power series' convergence radius by converting it to $\sum_{n=0}^{\infty}a_n (x-x_0)^n$, where $x_0$ is the centre of the convergence circle. The convergence radius $r$ can then be calculated with $\frac1{r}=\lim \limits_{n \to \infty}|\frac {a_{n+1}}{a_n}|$. But In the first series, $a_n=\sum_{k=0}^{n}\frac 1{k^3}$ is a series itself. How should I go about solving this exercise?

Also, does "examining their behaviour on the edge of their respective convergence circles" mean I should see if the series has a limit for $n \to (x_0-r)$ and $n \to (x_0+r)$?
 Thank you in advance for your help!

答:
The simplest is probably to use Hadamard's formula:

$$\frac 1 R=\limsup\sqrt[n]{a_n\vphantom{b}}=\limsup\sqrt[n]{a_n\vphantom{b}}.$$
Observe that
$$1<a_n <\zeta(3),\quad\text{ so }\quad 1 < \sqrt[n]{a_n} <\sqrt[n]{\zeta(3)}.$$
Can you take it from there?


问:
Determining the number of nontrivial families of increasing substrings of $\{1,...,k\}$
Let us say that we have the order $\{1,2,3,4,5,...,k\}$ of $k$ elements. Now, I want to determine in how many different ways can we pick a family of increasing substrings from $\{1,...,k\}$? The rules are the following:

We do not consider the trivial case of picking the entire $\{1,...,k\}$ as a whole.

We want to pick at least one increasing substring, but we can pick multiple different ones (such as if $k=5$ we can pick just $\{1,2\}$, but we may also pick $\{1,2\}$ and $\{3,4,5\}$, using the entire sequence and it would also be possible to pick something like $\{1,2\}$ and $\{3,4\}$).

The substrings may not overlap with one another (i.e. if for example $k=5$ we cannot have $\{1,2,3\}$ and $\{3,4\}$).

The increasing substrings must only consist of consecutive numbers (i.e. we cannot have anything like $\{1,3,4\}$).

A substring cannot be just one single number i.e. $\{1\}$.

So basically, we are curious about how many different combinations of these substrings we can have in total as families. An example would be, for the case of $k=4$ the following:

$\Big\{ \{1,2\}, \{1,2,3\}, \{2,3\}, \{3,4\}, \{2,3,4\}, \{\{1,2\}, \{3,4\}\} \Big\}$

so the number of families in this case would be $6$.

I have thought about it this way: if I have $\{1,...,k\}$ and enumerate them starting from only having one increasing substring such as $\{1,2\}$ or anything $\{1,...,j\}$ where $j \le k-1$. There are the total of $k-1$ of these.

Then we would enumerate any substrings of the length two on their own (minus 1 from the previous part), length three (minus 1 from the previous part), length $k-1$ (minus one from the previous part).

I believe how this is how we would do it for the case of considering only single substrings, but we also want to consider different combinations of multiple substrings, and I am not sure how to go about doing this.

答:
This is the same as asking for the number of subsets of $\{1, \cdots, k\}$ with evenly many elements (not including the empty set or the set $\{1,k\}$). Indeed, we just pair the elements in the subset sequentially to get the first and last elements of your substrings.  Thus, with $k=11$ the subset $\{3,6,7,11\}$ corresponds to the family $\{\{3,4,5,6\}, \{7,8,9,10,11\}\}$.

In general there are $2^{k-1}$ subsets with evenly many elements (see, e.g., this question).  As we wish to exclude the empty set and the set $\{1,k\}$ the answer is $$\boxed {S_k=2^{k-1}-2}$$

Note that  $S_3=4-2=2$ and $S_4=8-2=6$ in harmony with your examples.


问:
How to rewrite $\int\limits_{x = 0}^{ + \infty } {\frac{{\exp \left( { - ax - \frac{b}{x}} \right)}}{{1 + cx}}dx}$?
How to rewrite this improper integral in term of any special function $\int\limits_{x = 0}^{ + \infty } {\frac{{\exp \left( { - ax - \frac{b}{x}} \right)}}{{1 + cx}}dx}$ or at least written it as an infinite sum ?

This is a follow up post from the previous one
How to evaluate $\int\limits_0^\infty {\frac{{{e^{ - x - \frac{1}{x}}}}}{{1 + x}}dx}$ in form of special function?

That is $\int\limits_{x = 0}^{ + \infty } {\frac{{\exp \left( { - x - \frac{1}{x}} \right)}}{{1 + x}}}  = {K_0}\left( 2 \right)$

Where ${K_M}\left( x \right)$ denote the modified Bessel function of the second kind

I hope that I could generalized the old result !

Notice that $a,b,c$ are all positive number.

答:
Assuming $0<c<1$
$$I=\int\limits_{0}^{ \infty } {\frac{{\exp \left( { - ax - \frac{b}{x}} \right)}}{{1 + cx}}dx}=\sum_{n=0}^\infty (-1)^n c^n \int\limits_{0}^{ \infty } x^n\exp \left( { - ax - \frac{b}{x}} \right)\,dx$$
$$I=2\sum_{n=0}^\infty (-1)^n c^n a^{-\frac{n+1}{2} } b^{\frac{n+1}{2}} K_{-(n+1)}\left(2 \sqrt{a} \sqrt{b}\right) \qquad \Re(b)>0\land \Re(a)>0$$ that you could make nicer.


问:
Equivalence between two definitions of affine algebraic set.
In my algebraic geometry course, we have seen two definitions of an affine algebraic set. The first says that $X$ is an affine algebraic set if
$$X = \{x \in \mathbb{A}^n(k)~|~f_i(x) = 0, \forall i \in I\}$$
where $f_i \in k[X_1, \ldots, X_n]$ for all $i \in I$. The second one defines $X$ as a representable functor from $\text{Alg}_k$ to $\text{Set}$, i.e. there exists $A \in \text{Alg}_k$ such that
$$X = \text{Hom}_{\text{Alg}_k}(A, -).$$
Now the two definitions should match for $k$ algebrically closed and $A \in \text{Alg}_k$ finitely generated and reduced (= nilpotent-free).

There is a step in this equivalence that I don't understand: We take $X$ as in the first
definition, we consider the associated coordinate algebra
$$\mathcal{O}(X) = k[X_1, \ldots, X_n]/I(X)$$
and then we should have
$$\text{Hom}_{\text{Alg}_k}(\mathcal{O}(X), k) \cong X.$$
I don't really get this isomophism, does anyone know how to prove this or just a way to understand it intuitively ?

答:
Let $k$ be any field and let $B:=k[x_1,..,x_n]$ be the polynomial ring in $n$ variables. Let $I \subseteq B$ be an ideal with $A:=B/I$ and let $X:=Spec(A)$ and let $S:=Spec(k)$.

Let

R0. $X(k)$ be the set of $n$-tuples $(a_1,..,a_n)\in k^n$ with $f(a_1,..,a_n)=0$ for all polynomials $f \in I$.

R1. Let $Hom_{k-alg}(A,k)$  be the set of all maps of $k$-algebras $\phi: A \rightarrow k$.

Lemma. There is a one-to-one correspondence of sets $X(k) \cong Hom_{k-alg}(A,k)$.

If $a:=(a_1,..,a_n)\in X(k)$ we define a map $\phi_a: A\rightarrow k$ by letting $\phi_a(x_i):=a_i$. Conversely if $\phi: A\rightarrow k$ is a map of $k$-algebras it follows $\phi(x_i)=a_i \in k$ for all $i=1,..,n$. It follows $a:=(a_1,..,a_n)\in X(k)$. This is a one-to-one correspondence of sets.

People write $X(k)$ to denote the "set of $k$-rational points of $X$". A $k$-rational point is by definition a map of schemes

R2. $\phi^*: Spec(k) \rightarrow X$

which again corresponds 1-1 to a map of $k$-algebras $\phi:A \rightarrow k$. You do not need $k$ to be a field, you can let $k$ be any commutative unital ring. By the construction of the affine scheme $Spec(A)$ it follows

R3. $Hom_{Sch(S)}(S,X)\cong Hom_{k-alg}(A,k)$

Example 1. Let $I:=\{f_1,..,f_l\}$ be generated by a finite set of polynomials.
It follows a map of schemes $\phi^*: S\rightarrow X$ corresponds in a 1-1 way to a solution $(a_1,..,a_n)\in k^n$ of the system of polynomial equations

S1. $f_1(x_1,..,x_n)=0,..,f_l(x_1,..,x_n)=0$.

Hence the "functor of points" language converts the problem of studying sets of solutions to systems of polynomial equations into the problem of studying sets of maps $\phi^*: S \rightarrow X$.

Question: "Sorry I do not know this notation, what do you mean by Spec(A)?"

Answer: In chapter II in Hartshornes (or any book on algebraic geometry) book the ringed topological space $Spec(A)$ is introduced. Given any commutative unital ring $A$ you construct a topological space $X:=Spec(A)$ (it is the set of prime ideals in $A$ with the Zariski topology) and a structure sheaf $\mathcal{O}_X$. Given any map of unital commutative rings $\phi: A\rightarrow B$ (let $Y:=Spec(B)$) you get a map of locally ringed spaces

$(\phi^*, \phi^{\#}): (Y, \mathcal{O}_Y) \rightarrow (X, \mathcal{O}_X)$

with the property that the map of ringed topological spaces $(\phi^*, \phi^{\#})$ is uniquely determined by $\phi$. There is a 1-1 correspondence

R4. $Hom_{Sch}(Y,X) \cong Hom_{rings}(A,B)$.

This is Proposition II.2.3 in Hartshornes classical book "Algebraic geometry".
Hence you get a "geometric structure" $(X, \mathcal{O}_X)$ from any commutative unital ring $A$. And to study a map of schemes $\phi^*: S \rightarrow X$ is by the above argument equivalent to studying a solution to the system S1 from Example 1.

Why study solutions to systems of polynomial equations over rings that are not fields?

Example. In diophantine geometry one studies integral solutions to systems of polynomial equations. Given a set of polynomials $I:=\{f_1,..,f_l\}\in B:=\mathbb{Z}[x_1,..,x_n]$ we may construct the quotient ring $A:=B/I$. Let $S:=Spec(\mathbb{Z})$ and $X:=Spec(A)$. It follows there is a 1-1 correspondence of sets

R5 $Hom_{Sch}(S,X)\cong Hom_{rings}(A, \mathbb{Z})\cong X(\mathbb{Z})$.

For any scheme $X\in Sch(S)$ there is the "functor of points"

R6. $h_X: Sch(S) \rightarrow Sets$

where $Sch(S)$ is the category of schemes over $S$ and maps of schemes over $S$
and $Sets$ is the category of sets and maps of sets. The functor $h_X$ is defined by

R7. $h_X(T):=Hom_{Sch(S)}(T,X)$.

This gives an "embedding"

R8. $h: Sch(S) \rightarrow Funct(Sch(S), Sets)$.

Here $Funct(Sch(S), Sets)$ is the "category of functors" from $Sch(S)$ to $Sets$ with natural transformations as morphisms.

It is an "embedding" in the sense that any natural transformation $\eta:h_X \rightarrow h_Y$ of functors is induced by a unique morphism of schemes $f:X\rightarrow Y$, and two different morphisms $f,g:X\rightarrow Y$ give different natural transformations (this is the Yoneda lemma). With the functor $h$ we may view the category $Sch(S)$ as a "sub category" of $Funct(Sch(S), Sets)$. This viewpoint is what people use to define algebraic spaces, algebraic stacks and other "moduli spaces/parameter spaces" in algebraic geometry. As an example:

Let $E$ be a rank $d+1$ locally trivial $\mathcal{O}_S$-module and define the following functor:

R9. $F_{E}: Sch(S) \rightarrow Sets$

by $F_{E}(X,f):= \{\phi: f^*E^* \rightarrow L \rightarrow 0$ such that the sequence is exact $\}/\equiv$

where $L\in Pic(S)$ and two quotients $(L, \phi), (L', \phi')$ are equivalent iff there is an isomorphism $\psi: L \rightarrow L'$ such that the obvious diagram commutes. The functor $F_{E}$ is in the "category" $Funct(Sch(S), Sets)$, and it is in the "image" of the functor $h$: There is a scheme $\mathbb{P}(E^*)$ (the projective space bundle of $E$) and an isomorphism
of functors $h_{\mathbb{P}(E^*)}\cong F_{E}$. We say that the functor $F_E$ is "representable" and that "it is represented by $\mathbb{P}(E^*)"$. You may find this in Hartshorne, Chapter II.7, Proposition 7.12. Many functors
appearing in the study parameter spaces are not representable in this sense, and this leads to the theory of algebraic spaces and algebraic stacks.
"Many" "moduli problems" may be formulated as follows: Given a functor

$F: Sch(S) \rightarrow Sets.$

Question: Is there a scheme $f: X\rightarrow S$ with the property that $F(-) \cong h_X$ is an isomorphism of functors? Projective space bundles, grassmannian bundles, quot schemes etc are constructed using this language.
If $E$ is a locally trivial $\mathcal{O}_S$-module of rank $d+1$ you may construct the projective space bundle $\pi: \mathbb{P}(E^*)\rightarrow S$. There is the "tautological sequence"

T1. $\pi^*E^* \rightarrow \mathcal{O}(1) \rightarrow 0$.

If $S=k$ is a field it follows $E=V$ is a vector space of dimension $d+1$. If you dualize T1 you get an injection

T2. $\mathcal{O}(-1) \subseteq \pi^*E$

Given a $k$-rational point $x:Spec(k) \rightarrow \mathbb{P}(V^*)$
you get a line

T3. $\mathcal{O}(-1)(x) \subseteq V$ and this is a one to one correspondence.

This reflects that projective space is a parameter space: It parametrizes lines in $V$.

https://en.wikipedia.org/wiki/Moduli_space

There is the "Yoneda lemma":

https://en.wikipedia.org/wiki/Yoneda_lemma


问:
Why is $2\pi i \neq 0?$
We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$

Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?

答:
You have shown that $e^{2\pi i} = e^0$.  This does not imply $2\pi i = 0$, because $e^z$ is not injective.
You have to give up your intuition about real functions when you move them to the complex plane, because they change a lot.  $e^z$ is actually periodic for complex $z$.

ContentLicense: CC BY-SA 3.0

Title: Why is $2\pi i \neq 0?$


问:
Independent set of vectors in a dependent set.
If I have a dependent set of vectors, how do I know which of them are linearly independent between them, and which vectors exactly are dependent to the others?

For example: If I have 5×5 matrix of rank 4. Does that mean that if I take any 4 vectors out of the matrix they are linearly independent?If not, how do I know which ones are exactly independent and whis is the one dependent?

Thank you in advance and I apologise for not being very clear.

答:
If I have a dependent set of vectors, how do I know which of them are linearly independent between them, and which vectors exactly are dependent to the others?

Answer: Gauss's elimination.

Explanation: Suppose that you have the following set $S=\left\{v_{1},v_{2},v_{3},\ldots,v_{k}\right\}\subset V$ where $V$ is a vector space on field $F$. Assuming that we know that in the set $S$ there is a subset of $S$ that is dependent linearly and there is another subset of $S$ such that is independent linearly, so your question is: how to recognize which of the vectors is linearly dependent and which of the vectors is linearly independent? The answer is behind: Gaussian row reduction (row elimination).
Take the coordinates of the vectors in $S$, and then reduce by rows the matrix that contains the coordinate vectors in its rows, the rows that vanish were associated with the linearly dependent vectors and the rows that did not vanish were associated with the vectors linearly independent.

Example:  Let's to study the subset $$S=\left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} \right\}\subset \mathbb{R}^{3}$$

Now, in the set $S$ is clear that $\{(1,0,0),(0,1,0),(0,0,1)\}$ is independent linearly in $\mathbb{R}^{3}$ and $\{(1,1,1),(2,2,2)\}$ is dependent linearly in $\mathbb{R}^{3}$. But let's to use the elimination method for check it:

Let $$A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 1\\ 2 & 2 & 2 \end{pmatrix}$$
for Gauss's elimination we can see that
$$A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \color{blue}{1} & \color{blue}{1} & \color{blue}{1}\\ \color{red}{2} & \color{red}{2} & \color{red}{2} \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \color{blue}{0} & \color{blue}{0} & \color{blue}{0}\\ \color{red}{0} & \color{red}{0} & \color{red}{0} \end{pmatrix} $$

since that the row $\color{blue}{4}$ and $\color{red}{5}$ is associated for the vectors $(1,1,1)$ and $(2,2,2)$,  then by the Gauss's elimination we can conclude that the set $\{(1,1,1),(2,2,2)\}$ is dependent linearly in $\mathbb{R}^{3}$ and that the set $\{(1,0,0),(0,1,0),(0,0,1)\}$ is independent linearly in $\mathbb{R}^{3}$.


问:
Argument of Feynman for equivalence of dot product definitions
The following argument from the Feynman Lectures on Physics (Vol I, Lecture 11), which relates to the equivalence of the algebraic and geometric definitions, does not particularly convince me.

Also, there is a simple geometrical way to calculate $\vec{a} \cdot \vec{b}$, without
having to calculate the components of $\vec{a}$ and $\vec{b}$: $\vec{a} \cdot \vec{b}$ is the product of
the length of $\vec{a}$ and the length of $\vec{b}$ times the cosine of the angle
between them. Why? Suppose that we choose a special coordinate system
in which the x-axis lies along $\vec{a}$; in those circumstances, the only
component of $\vec{a}$ that will be there is $a_x$, which is of course the whole
length of $\vec{a}$. Thus Eq. (11.19) reduces to $a \cdot b = a_x b_x$ for this case, and
this is the length of $\vec{a}$ times the component of $\vec{b}$ in the direction of
$\vec{a}$, that is, $b \cos \theta$: $a \cdot b = a b \cos \theta$. Therefore, in that special coordinate
system, we have proved that $\vec{a} \cdot \vec{b}$ is the length of $\vec{a}$ times the length of
$\vec{b}$ times $\cos \theta$. But if it is true in one coordinate system, it is true
in all, because $\vec{a} \cdot \vec{b}$ is independent of the coordinate system; that is
our argument.

In fact, most of this argument seems just fine, but it seems like Feynman is casually asserting a priori that the dot product should be independent of the coordinate system. This is something I do not like, since I can't see an obvious justification for it. (Indeed, if by "coordinate system" he means basis, then there are clearly bases for which this is not true, e.g., ${2\hat{i}, 2\hat{j}, 2\hat{k}}$.)

Could someone who is better at reading between the lines of Feynman please clarify this for me?

答:
If I read correctly that you're looking for a justification that the two computations are equivalent, I came up with a way to demonstrate this a few months ago. Myself, I always found the geometric $\vec{a} \cdot \vec{b} = ||\vec{a}|| \cos(θ) ||\vec{b}||$ computation to be the more intuitive, so I'll start with that definition and justify the element-based definition. This won't require anything more than basic trig and some geometric intuitions.

Say we have 2 vectors, $ \vec{u} $ and $ \vec{v} $, in $ \mathbb{R}^n$, and let's assume that these vectors are non-collinear-- one isn't a scaled up version of the other (if they are collinear, then cos(θ) = 1 which simplifies the problem considerably). Explicitly,
$$\vec{u}=<u_1,u_2,u_3,\cdots,u_n>$$
$$\vec{v}=<v_1,v_2,v_3,\cdots,v_n>$$
If they aren't collinear, and if their tails lie at the origin of our n-dimensional space, then we can use their positions two define two lines: one line going between  the origin and $\vec{u}$, and another line going through the origin and $\vec{v}$. The great thing about this setup is that now we're dealing with 3 points in space. It doesn't matter how many dimensions those points may exist in according to our axes, because 3 points is enough to define a unique plane in that space. We don't know how to label those points in that plane, but we know those points exist in it. Turning to GeoGebra,

I've labeled the coordinate points in context of our n-dimensional space-- we're not using the 2D coordinates of the plane itself. Now, we know we want to connect this element computation to geometry. To do that, I've labeled θ, the angle between the two vectors in the plane that contains them, and I've defined a vector $$\vec{w} = \vec{v}-\vec{u}$$
$$\vec{w} = <v_1 - u_1, v_2 - u_2, v_3 - u_3, \cdots, v_n - u_n>$$
We notice that this diagram draws out a triangle, and, if we assume we know θ as well as our two vectors, we get the triangle

And now we do some math.

It's not guranteed that our vectors will produce a right triangle, so to solve this thing we'll need the Law of Cosines, as well as the magnitudes of the sides of our triangle.
$$\text{Law of Cosines:} \: \: c^2 = a^2 +b^2 - 2ab\cos(\angle C)$$
$$ ||\vec{u}|| = \sqrt{u^2_1 + u^2_2 + u^2_3 + \cdots + u^2_n}$$
$$ ||\vec{v}|| = \sqrt{v^2_1 + v^2_2 + v^2_3 + \cdots + v^2_n}$$
$$ ||\vec{w}|| = \sqrt{(v_1 - u_1)^2 + (v_2 - u_2)^2 + (v_3 - u_3)^2 + \cdots + (v_n - u_n)^2}$$
If our side $a$ is represented by $\vec{u}$, and side $b$ is represented by $\vec{v}$, then side $c$ and $\angle C$ are given by $\vec{w}$ and $θ$. Using these values in the Law of Cosines formula and solving this for $\cos(θ)$ gives
$$\cos(\angle C) = -\frac{c^2 - a^2 - b^2}{2ab}$$
$$\cos(θ) = -\frac{||\vec{w}||^2 - ||\vec{u}||^2 - ||\vec{v}||^2}{2||\vec{u}|||\vec{v}||}$$
$$\cos(θ) = -\frac{[(v_1 - u_1)^2 + (v_2 - u_2)^2 + \cdots + (v_n - u_n)^2] - [u^2_1 + u^2_2 + \cdots + u^2_n] - [v^2_1 + v^2_2 + \cdots + v^2_n]}{2||\vec{u}|||\vec{v}||}$$
Expanding out our terms for $||\vec{w}||^2$,

$$\cos(θ) = -\frac{
[v_1^2 - 2 u_1 v_1 + u_1^2 + v_2^2 - 2 u_2 v_2 + u_2^2 + \cdots + v_n^2 - 2 u_n v_n + u_n^2]
 - [u^2_1 + u^2_2 + \cdots + u^2_n] - [v^2_1 + v^2_2 + \cdots + v^2_n]}{2||\vec{u}|||\vec{v}||}$$
It is at this point we notice that every $u_i^2$ and $v_i^2$ in the numerator has a corresponding $-u_i^2$ and $-v_i^2$, allowing us to reduce all those terms to $0$. Doing this, along with distributing the negatives across, gives us
$$\cos(θ) = \frac{2 u_1 v_1 + 2 u_2 v_2 + \cdots + 2 u_n v_n}{2||\vec{u}|||\vec{v}||}$$
$$\cos(θ) = \frac{ u_1 v_1 +  u_2 v_2 + \cdots +  u_n v_n}{||\vec{u}|||\vec{v}||}$$
Tying all of this back to our given definition for the dot product
$$\vec{u} \cdot \vec{v} = ||\vec{u}||\cos(θ)||\vec{v}||$$
$$\vec{u} \cdot \vec{v} = ||\vec{u}||||\vec{v}||\frac{ u_1 v_1 +  u_2 v_2 + \cdots +  u_n v_n}{||\vec{u}|||\vec{v}||}$$
$$\vec{u} \cdot \vec{v} = u_1 v_1 +  u_2 v_2 + \cdots +  u_n v_n$$
Hopefully that gives an intuitive path for how we would go about connecting those two computations.


问:
Find the number of positive integers $n < 2007$ such that $\Big[\frac{n}{2}\Big]$ + $\Big[\frac{n}{3}\Big]$ + $\Big[\frac{n}{6}\Big] = n$ .

Find the number of positive integers $n < 2007$ such that $\Big[\frac{n}{2}\Big]$ + $\Big[\frac{n}{3}\Big]$ + $\Big[\frac{n}{6}\Big] = n$ . ($[]$ denotes the floor function)

What I Tried: Till now I am not able to find any good ideas on how to attack this problem. For example the LHS is an integer, and so will be the RHS, but it is of no use as it is alreadt mentioned we need to find positive integers $n$ .

Next, assume $n = 6k + a$ for $k \in \mathbb{N}$ and $0 \leq a < 6$ .
Then the equation becomes :-
$$3k + \Big[\frac{a}{2}\Big] + 2k + \Big[\frac{a}{3}\Big] + k = 6k + a$$
$$\rightarrow 6k + \Big[\frac{a}{2}\Big] + \Big[\frac{a}{3}\Big] = 6k + a$$
$$\rightarrow \Big[\frac{a}{2}\Big] + \Big[\frac{a}{3}\Big] = a $$

Can anyone help me after this?

答:
Note that the floor function satisfies the inequality
$$[x] \le x$$ and equality holds if and only if $x$ is an integer.
Now
$$n = \left[ \frac{n}{2} \right]+\left[ \frac{n}{3} \right]+\left[ \frac{n}{6} \right] \le \frac{n}{2} + \frac{n}{3} + \frac{n}{6} =n$$
This shows that the three fractions must be integers.

Can you conclude from here?


答:
The $\log$ function is multi-valued on $\mathbb{C}^*$ (you can however choose a "branch" of it; see Wikipedia).  At any rate, just because
$$e^{2\pi i}=e^0$$
does not imply that $2\pi i=0$.</p>

Title: Why is $2\pi i \neq 0?$

Body:
We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$

Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?


问:
Probability of $p(x_1<x_2<\max(x_3,x_4))$
If $x_1,x_2,x_3$ and $x_4$ are independent identically distributed  random variable with continuous distribution .what is the probability of

$$P=p(x_1 < x_2< \max(x_3,x_4))$$
I know the exhaustive number of cases are $p(x_1<x_2<x_3)+p(x_1<x_2<x_4)$ and  every $x$ are iid so both have same probabiltiy so is $2p(x_1<x_2<x_3)$ is right? But  i get the $p(x_1<x_2<x_3) = 1/6$ and then my answer is $1/3$ but the answer is $1/6.$ Can we write $P$ as $p(x_1<x_2<x_3<x_4)+p(x_1<x_2<x_4<x_3)$?

Body:
$$p(x_1\lt x_2\lt\max(x_3,x_4))=p(x_1\lt x_2\lt x_3\text{ OR }x_1\lt x_2\lt x_4)$$
$$=p(x_1\lt x_2\lt x_3)+p(x_1\lt x_2\lt x_4)-p(x_1\lt x_2\lt x_3\text{ AND }x_1\lt x_2\lt x_4)$$
$$=p(x_1\lt x_2\lt x_3)+p(x_1\lt x_2\lt x_4)-p(x_1\lt x_2\lt x_3\lt x_4\text{ OR }x_1\lt x_2\lt x_4\lt x_3)$$
$$=\frac1{3!}+\frac1{3!}-\frac1{4!}-\frac1{4!}=\frac14.$$


问:
Let's assume 1 kilogram is equal to 2.2 pounds. But I got: 1 : 2.2 = lb : kg
Let's assume 1 kilogram is equal to 2.2 pounds.

$1 kg = 2.2 lb$

So we get:

$1/2.2 = lb/kg$

Which means:

$1 : 2.2 = lb : kg$

So if we convert 5 kg to lb and y stands for the pound number, we get:

$1 : 2.2 = ylb : 5kg$

$5kg= y*2.2*lb$

$5(2.2lb)= y*2.2*lb$

$y = 5$

It is really wierd because we all know that it should have been

$kg : lb = 1 : 2.2$

Is there a lack of thinking from me?

答:
Well, if $y$ is the number such that $y\,lb=5\,kg$, then you'd have $y\, lb : 5\,kg = 1:1$, no? Because they are supposed to be the same mass. What you should have is that $y\times lb = 5\times kg$, i.e. $$ y:5=kg:lb=2.2:1$$

Or, as some would say in a rather convoluted way, the numerical values are inversely proportional to units of measure.


答:
$1 : 2.2 = lb : kg\tag{1}$

So if we convert 5 kg to lb and y stands for the pound number, we get:

$1 : 2.2 = ylb : 5kg$

This is completely wrong, like saying that $\frac{1}{2}=\frac{5}{2}$

If you want to know how many lb are 5 kg you just multiply the first line by 5

1kg = 2.2lb so 5kg are 11lb

Title: Let's assume 1 kilogram is equal to 2.2 pounds. But I got: 1 : 2.2 = lb : kg

Body:
Let's assume 1 kilogram is equal to 2.2 pounds.

$1 kg = 2.2 lb$

So we get:

$1/2.2 = lb/kg$

Which means:

$1 : 2.2 = lb : kg$

So if we convert 5 kg to lb and y stands for the pound number, we get:

$1 : 2.2 = ylb : 5kg$

$5kg= y*2.2*lb$

$5(2.2lb)= y*2.2*lb$

$y = 5$

It is really wierd because we all know that it should have been

$kg : lb = 1 : 2.2$

Is there a lack of thinking from me?


问:
Convergence in distribution for random functions
I am a bit confused about how to fix some feedback I got.

For a paper I am writing for a class, I used the following to explain why a certain random walk converges to Brownian motion, the following is a direct snip from my paper:

Definition (Convergence in distribution)

A sequence of random variables $X_1,X_2,\dots,$ converges in distribution to a random variable $X$ if
\begin{equation}
    \lim _{n \rightarrow \infty} F_{X_{n}}(x)=F_{X}(x)
\end{equation}
at all points $x$ where $F_{X}(x)$ is continuous. Where $F_{X_{n}}$ and $F_{X}$ are the cumulative distribution functions of random variables $X_n$ and $X$ respectively.

This scaled random walk on $[0,1]$ then converges to a one-dimensional standard Brownian motion by the following theorem:

Theorem (Donsker's Theorem )

Let $\{X_i , i \geq 1 \}$ be a sequence of real-valued i.i.d. random variables such that $\mathbb{E}(X_1) = 0$, and $\operatorname{Var}(X_1) = 1$. Let $\{ S_n , n \geq 0 \}$ be the one-dimensional random walk based on $\{X_i , i \geq 1 \}$ and define $\left\{W_{n}(t), 0 \leq t \leq 1\right\}$ as $W_n(0) = 0$ and
\begin{equation}
W_{n}(t)=\frac{1}{\sqrt{n}} S_{\lfloor n t \rfloor }+(n t-\lfloor n t \rfloor) \frac{1}{\sqrt{n}} X_{\lfloor
n t \rfloor
+1}, \quad t \in(0,1], n \geq 1 \label{Donsker}
\end{equation}
Then, $W_{n}$ converges in distribution to $\mathbb{W}$ as $n \rightarrow \infty$, where $\mathbb{W}$ denotes the one-dimensional standard Brownian motion.

The feedback I received said that I cannot directly say $W_{n}$ converges in distribution to $\mathbb{W}$, as $W_{n}$ is not one random variable but a random function. I have tried to find a similar definition for random functions however I havent had much luck. Does anyone know a reasonable way to fix this?

答:
Here convergence of $W_n$ to $W$ in distribution means convergence of the induced measures on the space $C[0,1]$. $W_n$ induces a measure $\mu_n$ on this space and $W$ induces the Wiener measure $\mu$. $\mu_n \to \mu$ means $\int Fd\mu_n \to \int F d\mu$ for every bounded continuous function $F:C[0,1] \to \mathbb R$.

We refer to $\mu_n$ and $\mu$ as the distributions of the processes $W_n$ and $W$.


问:
Does the set of all formal Laurent series form an ordered field?
Let $R$ be an ordered field. Consider the set $R[[x]]:=\{\sum_\limits{n \in \mathbb{Z}}a_nx^n:a_n \in R\}$

Addition:
$\sum_\limits{n \in \mathbb{Z}}a_nx^n+\sum_\limits{n \in \mathbb{Z}}b_nx^n:=
\sum_\limits{n \in \mathbb{Z}}(a_n+b_n)x^n$

Multiplication:
$\sum_\limits{n \in \mathbb{Z}}a_nx^n \times \sum_\limits{n \in \mathbb{Z}}b_nx^n:=
\sum_\limits{n \in \mathbb{Z}}(a_nb_n)x^n$

Does these operations on $R[[x]]$ makes it a field?

Have I correctly defined the multiplication?

Is the notation $R[[x]]$, i.e does the notation $R[[x]]$ have some other meaning?

If yes, is there any standard notation for this set?

What order relation can be defined on $R[[x]]$ in order to turn it into an ordered field?

My motive is to show that $R[[x]]$ is Cauchy Complete but is not Order Complete(i.e lub property)

Edit: As said by @ the multiplication defined above doesn't even make it an integral domain.

So I am changing the set to $R[[x]]:=\{\sum_\limits{k=-n}a_kx^k: a_k \in R, n \in \mathbb{N} \}$  and multiplication as defined in the answers.

答:
No, it's not a field with the multiplication you defined! The multiplicative identity is $\sum z^n$, hence $z$ is a non-zero element with no multiplicative inverse.

Where did this come from? Offhand I don't see how to define a multiplication that makes $R[[x]]$ into a field. The natural multiplication for formal power series is $$\left(\sum a_nx^n\right)\left(\sum b_n x^n\right)=\sum_n\left(\sum_{j+k=n}a_jb_k\right)x^n,$$but of course that doesn't converge for formal Laurent series...


问:
Why does Spivak define $f$ to be integrable if $\sum_{\phi \in \Phi} \int_A \phi \cdot |f|$ converges
When Spivak defines integration with partitions of unity on page 65 in Calculus on Manifolds, he states:

"We define $f$ to be integrable (in the extended sense) if $\sum_{\phi \in \Phi} \int_A \phi \cdot |f|$ converges. This implies convergence of $\sum_{\phi \in \Phi} |\int_A \phi \cdot f|$ and hence absolute convergence of $\sum_{\phi \in \Phi} \int_A \phi \cdot f$, which define to be $\int_A f$."

He seems to use a stricter condition than necessary, specifically that $\sum_{\phi \in \Phi} \int_A \phi \cdot |f|$ converges. Why does he not broaden the definition of $f$ to be integrable if $\sum_{\phi \in \Phi} |\int_A \phi \cdot f|$ converges.

In the latter case, we still have absolute convergence, and since $\int_A \phi \cdot |f| \text{ converges} \implies |\int_A \phi \cdot f| \text{ converges} $, but not necessarily the other way around, we have a broader definition.

答: Partitions of unity are a technical tool that allow us to define integration in a more general setting. As such, we would like $\int_Af=\sum_{\phi\in\Phi}\int_A\phi f$ to depend only on $A$ on $f$, but not on the chosen partition of unity $\Phi$. Indeed, this is the case and the content of Theorem 3-12(1). Try mimicking the proof of this theorem if you only suppose convergence of $\sum_{\phi\in\Phi}\vert\int_A\phi f\vert$ instead of $\sum_{\phi\in\Phi}\int_A\phi\vert f\vert$. You will not be able to do this. Spivak remarks, right between the definition and the theorem, "These definitions do not depend on $\mathcal{O}$ or $\Phi$ (but see Problem 3-38)". Problem 3-38 gives a scenario in which you can find two partitions of unity $\Phi,\Psi$ such that $\sum_{\phi\in\Phi}\int_{\mathbb{R}}\phi f$ and $\sum_{\psi\in\Psi}\int_{\mathbb{R}}\psi f$ both converge absolutely, but to different values. It is instructive to think this through.


答:
I think that using formulae to calculate the radius of convergence is misguided, and that the comparison test is in general a better way to proceed.

Let $a_n:=\sum_{k=1}^{n}\frac{1}{k^3}$, and consider the series $\sum_n a_n x^n$ whose radius of convergence is $R$.

Case (i) $x>1$. We have that $a_n x^n >x^n$, and $\sum x^n$ is divergent; so by comparison  $\sum a_n  x^n$ is also divergent. Hence $R\leqslant 1$.

Case (ii) $x<1$. Note that $\sum_{k=1}^{\infty}\frac{1}{k^3}$ is convergent, to $S$ say. Then the sequence $a_n$ is monotone increasing and bounded by $S$. Hence for each $x$ we have that $a_n x^n < Sx^n$. Now for such $x$ the series $\sum Sx^n$ is convergent, and so by comparison $\sum a_n  x^n$ is also convergent. Hence $R\geqslant 1$.

That is, the radius of convergence is $1$.

It remains to discuss convergence on the circle of convergence. As $|a_n x^n|=a_n$, and $a_n\not\to 0$, the series can't converge at these points.

Your second series can also be done by comparison.

Title: What should I do when the factor $a_n$ in a power series is a series itself?

Body:
We were given the following exercise:

calculate the convergence radius of the following series and examine their behaviour on the edge of their respective convergence circles:
$$\sum_{n=0}^{\infty}(1+\frac 1{8}+\frac 1{27}+\ldots+\frac 1{n^{3}})x^n$$
$$\sum \frac {x^{n^2}}{2^n}$$

now correct me if I'm wrong, but you can calculate a power series' convergence radius by converting it to $\sum_{n=0}^{\infty}a_n (x-x_0)^n$, where $x_0$ is the centre of the convergence circle. The convergence radius $r$ can then be calculated with $\frac1{r}=\lim \limits_{n \to \infty}|\frac {a_{n+1}}{a_n}|$. But In the first series, $a_n=\sum_{k=0}^{n}\frac 1{k^3}$ is a series itself. How should I go about solving this exercise?

Also, does "examining their behaviour on the edge of their respective convergence circles" mean I should see if the series has a limit for $n \to (x_0-r)$ and $n \to (x_0+r)$?
 Thank you in advance for your help!


问:
Deriving this Maclaurin expansion approximation of an even function at the origin $r=0$
For an even function $f(r)$ that is non-zero at $r=0$ and symmetric around $r=0$, i.e. $f(r)=f(-r)$, I want to be able to approximate the expression

$$
\frac{1}{r}\frac{\partial f}{\partial r} \Bigg\vert_{r=0}. \tag{1}
$$

However, since I want the approximation at $r=0$, the problem is that the $1/r$ factor makes it diverge. (The purpose of wanting to do this is ultimately to numerically solve a PDE in radial coordinates using finite differences).

I have seen several times (see for example Eq. 12 of this paper or p. 360 of this paper) that the following approximation should hold at the origin, in order to get around the singularity problem
$$
\frac{1}{r}\frac{\partial f}{\partial r}\Bigg\vert_{r=0}\approx \frac{\partial^2f}{\partial r^2}\Bigg\vert_{r=0},  \tag{2}
$$
and I would like to see where it comes from.

If I try to derive this approximation by Maclaurin expansion in some small $\delta r$ about the point $r=0$, I get the following up to second order:
$$
f(\delta r) \approx f(0) + \delta r \frac{\partial f}{\partial r}  \Bigg\vert_{r=0} + \frac{1}{2} \delta r^2\frac{\partial^2 f}{\partial r^2}  \Bigg\vert_{r=0} + \,\,\,\,...   \tag{3}
$$
Rearranging, this becomes
$$
\begin{align}
\frac{\partial^2 f}{\partial r^2}  \Bigg\vert_{r=0} &\approx \frac{2}{\delta r^2}\Bigg( f(\delta r) -  f(0) - \delta r \frac{\partial f}{\partial r}  \Bigg\vert_{r=0} \Bigg)  \tag{4}\\
&= \frac{2}{\delta r}\Bigg( \frac{f(\delta r) -  f(0)}{\delta r} -  \frac{\partial f}{\partial r}  \Bigg\vert_{r=0} \Bigg) \tag{5}
\end{align}
$$
Now I know that the first derivative is defined as
$$
\frac{\partial f}{\partial r}\Bigg\vert_{r=0}  = \lim_{\delta r\rightarrow 0}\frac{f(\delta r)-f(0)}{\delta r}  \tag{6}
$$
so for small $\delta r$ the equation should become
$$
\begin{align}
\frac{\partial^2 f}{\partial r^2}  \Bigg\vert_{r=0} &= \frac{2}{\delta r}\Bigg( \frac{\partial f}{\partial r}\Bigg\vert_{r=0}-  \frac{\partial f}{\partial r}  \Bigg\vert_{r=0} \Bigg)  \tag{7}\\
&= 0,
\end{align}
$$
which is not the same as the approximation given in Eq. (2). Where have I gone wrong here? How can I arrive successfully at Eq. (2)?

Thanks!

答:
The derivative of an even function is odd and is zero at the origin.

By Taylor on $f'$,

$$f'(r)=rf''(0)+o(r)$$

and

$$\lim_{r\to 0}\frac{f'(r)}r=f''(0).$$

Notice that approximation is understood as

$$\left.\left(\frac1r\frac{\partial f}{\partial r}\right)\right|_{r=0}$$ and not

$$\frac1r\left.\left(\frac{\partial f}{\partial r}\right|_{r=0}\right).$$


问:
Linear Algebra Terminology Question
What does the term $\mathcal{D}^{(3)}(\mathbb{R})$ mean?

I encountered the term in the context of this question in a class I am taking:

Let $W = \{f \in \mathcal{D}^{(3)}(\mathbb{R}) | f''' − 2f'' + 3f' − 4f = 0\}$. Is $W$ a subspace of $\mathcal{D}^{(3)}(\mathbb{R})$?

I stress that this is an assigned question related to a course I am taking, so I am not seeking a solution to this question, only an explanation regarding what the term noted above means (the set of which $f$ is a member).  I have never encountered it before.

答: $D^{(3)}(\Bbb{R})\equiv D^{(3)}(\Bbb{R},\Bbb{R})$ probably refers to the set of functions $f:\Bbb{R}\to\Bbb{R}$ such that $f$ is thrice differentiable on $\Bbb{R}$, i.e $f',f'',f'''$ all exist (you can easily show that $D^{(3)}(\Bbb{R},\Bbb{R})$ is a vector subspace of $\mathcal{F}(\Bbb{R},\Bbb{R})$, the set of all functions $\Bbb{R}\to\Bbb{R}$).

Title: Linear Algebra Terminology Question


答:
$\mathbb R[[X]]$ is usually reserved for formal power series. Formal Laurent series are denoted $\mathbb R((X))$. However, they look different than your version. The main difference: you defined multiplication componentwise, essentially making your ring the same as $\mathbb Z^{\mathbb R}$, the ring of real valued maps on the integers (each component $a_nx^n$ of the series corresponds to the fact that $n\mapsto a_n$). This ring is not a field. It's not even an integral domain, since $(X)\cdot(X^2)=0$, for instance, so we have zero-divisors.

The usual way to make the Laurent series into a field (assuming we are considering coefficients from a field, which we do here) is to stipulate that they may only have a finite amount of negative powers and to define multiplication as

$$\left(\sum_{k=-n}^\infty a_k X^k\right)\left(\sum_{k=-m}^\infty b_k X^k\right):=\sum_{k=-(n+m)}^\infty \sum_{i+j=k}a_ib_j X^k.$$

The reason for the finite amount of negative powers is to make the second sum finite, since infinite sums are not well defined in general.

As for a suitable ordering, you can define a positive cone $P\subset\mathbb R((X))$ as the set of all formal Laurent series whose leading coefficient (meaning the one with the smalles index) is nonnegative. With the ordering $f\geq g:\Leftrightarrow f-g\in P$ you get an ordered field.

Title: Does the set of all formal Laurent series form an ordered field?

Body:
Let $R$ be an ordered field. Consider the set $R[[x]]:=\{\sum_\limits{n \in \mathbb{Z}}a_nx^n:a_n \in R\}$

Addition:
$\sum_\limits{n \in \mathbb{Z}}a_nx^n+\sum_\limits{n \in \mathbb{Z}}b_nx^n:=
\sum_\limits{n \in \mathbb{Z}}(a_n+b_n)x^n$

Multiplication:
$\sum_\limits{n \in \mathbb{Z}}a_nx^n \times \sum_\limits{n \in \mathbb{Z}}b_nx^n:=
\sum_\limits{n \in \mathbb{Z}}(a_nb_n)x^n$

Does these operations on $R[[x]]$ makes it a field?

Have I correctly defined the multiplication?

Is the notation $R[[x]]$, i.e does the notation $R[[x]]$ have some other meaning?

If yes, is there any standard notation for this set?

What order relation can be defined on $R[[x]]$ in order to turn it into an ordered field?

My motive is to show that $R[[x]]$ is Cauchy Complete but is not Order Complete(i.e lub property)

Edit: As said by @ the multiplication defined above doesn't even make it an integral domain.

So I am changing the set to $R[[x]]:=\{\sum_\limits{k=-n}a_kx^k: a_k \in R, n \in \mathbb{N} \}$  and multiplication as defined in the answers.


问:
Show that a linear transformation exists where a inversible transformation is given
I've been trying to solve this issue:

Let $V$ be a finite vector space, and let $S:V\rightarrow V$ be an inversible linear transformation.

Show that a linear transformation, $T:V\rightarrow V$, exists where: $ker TS = \{0\}$ but $ker T \ne \{0\}$

Given the information, we know that $S$ is isomorphic, and there exists $S^{-1}$ which for each $v \in V$ makes $SS^{-1}(v) = v$ and $S^{-1}S(v) = v$.

And also we know that the kernel of the $S$, it being isomorphic. So, $ker S = \{0\}$.

How can I show the kernel of $ker TS \ne \{0\}$?

Thank you for answering!

答: It's not true: if $v\in\ker T$ nonzero then $S^{-1}v\ne 0$ and $S^{-1}v\in\ker TS$.


问:
Is there a reasonably accurate and easy way to approximate lbs-stones or kg-stones (and vice versa)?
For example, when I need to talk with my American friends about body weight:

$\mathrm{KG}\times2+10\%$

($100\times2=200$, $200+10\%=220$... Pretty accurate and easy to do mentally, both ways)

However, when I need to talk with my British friends about body weight, how do I approximate from KG to Stones?

Additionally, if I have a similar formula for Stones-lbs, that will be useful too!

I mean, I know a Stone is 14 lbs, but how to approximate 247lbs in Stones without pulling out a calculator?

For your reference, a Stone is about 6.35 KG

I am thinking that a "good, practical" approximation follows these rules:

5% accuracy is kinda bad... Imagine weighing 100kg and wanting to lose 5kg - the error alone is making discussion about it difficult! I am thinking up to 2% error

Maximum of 3 steps

Only addition, subtraction, multiplication and division allowed (% also allowed)

Multiplication only by "easy" factors: 2, 3, 4, 5, 10

Division only by "easy divisors": 2, 10

Only "easy" percentages like: 1%, 10%...

答:
70 kg is extremely close to 11 stone (about $0.2\%$ out). Now that perhaps doesn't give a sufficiently easy conversion because of the difficulty of dividing by $7$, but for weights of people (which is basically the only time anyone uses stones) you will generally be able to do well by taking the nearest multiple of 70 kg, then approximating the remainder in some cruder way (such as $\mathrm{kg}\times3/20$). For example taking 100 kg, $100=70+30$ gives an approximation of $11+3\times30/20=15.5$ stone. The actual value is $15.72$ stone, so we are about $1.4\%$ out.

(For people with weight significantly below 70 kg you can do better by approximating the difference from 35 kg $\approx$ 5.5 stone.)


问:
Find the number of real values of $x$ such that $x^2 + 10000[x] = 10000x$ ($[]$ is the floor function)

Find the number of real values of $x$ such that $x^2 + 10000[x] = 10000x$ ($[]$ is the floor function)

What I Tried: If $x$ is an integer then :-
$$x^2 = 0$$
$$\rightarrow x = 0$$
So $x = 0$ is the only integer solution, the problem with me is to look for real solutions. I have no clear idea on how to start doing it. One thing I did was let $[x] = x - \{x\}$ . We have :-
$$x^2 + 10000(x - \{x\}) = 10000x$$
$$\rightarrow x^2 - 10000\{x\} = 0$$
$$\rightarrow x^2 = 10000\{x\}$$
Can anyone help me here?

答:
Let

$$x = n + r, \; n \in \mathbb{Z}, \; 0 \le r \lt 1 \tag{1}\label{eq1A}$$

Your equation then becomes

$$\begin{equation}\begin{aligned}
x^2 + 10000\lfloor x\rfloor & = 10000x \\
(n + r)^2 + 10000n & = 10000(n + r) \\
n^2 + 2nr + r^2 - 10000r & = 0
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Treat the above equation as a quadratic in $n$ and use the quadratic formula to get

$$\begin{equation}\begin{aligned}
n & = \frac{-2r \pm \sqrt{(2r)^2 - 4(r^2 - 10000r)}}{2} \\
& = \frac{-2r \pm \sqrt{40000r}}{2} \\
& = -r \pm 100\sqrt{r}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Note $-r - 100\sqrt{r}$ is a strictly decreasing function, going from $0$ down to just greater than $-1 - 100 = -101$, i.e., $(-101, 0]$, so the integer values are $-100 \le n \le 0$. With $-r + 100\sqrt{r}$, it's a strictly increasing function, going from $0$ to just less than $-1 + 100 = 99$, i.e., $[0, 99)$, so the integer values are $0 \le n \le 98$. Altogether, the integer values are $-100 \le n \le 98$, with just one value of $r$ for each integer $n$, giving a total of $199$ solutions.


问:
Probability related to min and max of Brownian motion
Given the Brownian motion $X(t)$ and $s$ with value greater than $0$.

I want to ask about the probability:

\begin{align*}
\Pr(\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s) .
\end{align*}

Intuitively, I believe $\Pr(\min_{0 \leq u \leq t} X(u) > 0)$ and $\Pr(\max_{0 \leq u \leq t} X(u) > s)$ are independent when $s > 0$. So
\begin{align*}
\Pr(\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s) =  \Pr(\min_{0 \leq u \leq t} X(u) > 0)\Pr(\max_{0 \leq u \leq t} X(u) > s).
\end{align*}
But I just could not prove it logically.

Can someone give me a hint? I really appreciate it. Thank you very much.

答:
$$
\mathsf{P}\!\left(\min_{0\le u\le t}X_u>0,\max_{0\le u\le t}X_u>s\right)\le \mathsf{P}\!\left(\min_{0\le u\le t}X_u>0\right)=0.
$$


答:
It's like saying that, because $\sin{\pi} = \sin{0}$, that $\pi = 0$. Not all functions have perfect inverses; sin being one of them. In the complex numbers, $e^z$ doesn't either.

You're implicitly going: $e^{2\pi i} = e^0 \implies \ln{e^{2\pi i}} = \ln{e^0} \implies 2 \pi i = 0$. The error here is that $\forall x \in \mathbb{C} \ \ln{e^x} = x$ is not true! $\ln$ isn't even a function, just like the naive version of $\arcsin(x)$. You have to make a choice of range, which is usually $\Im(x) \in (-\pi, \pi]$.

Title: Why is $2\pi i \neq 0?$


问:
Prove that semiconvex function can be approximated by smooth functions with same semiconvexity constant
On page 60 of the user's guide to viscosity solutions, proof of Lemma A.3, it is stated that given $\varphi$,  a semiconvex function, we can find a smooth approximation $\varphi_\epsilon$ with the same semiconvexity constant as $\varphi$. Is there a reference to this fact? I tried proving it by taking $\varphi_\epsilon$ to be the convolution of $\varphi$ with an approximation to the identity, but it does not seem straightforward to prove that this satisfies the same semiconvexity constant as $\varphi$.

答:
The definition of semi-convexity being used here is $\varphi$ is semi-convex with constant $\Lambda > 0$ if $\varphi = \psi + \eta$, where $\psi$ is a convex function and $\eta$ a smooth function with $D^{2} \eta \geq - \Lambda \text{Id}$ (pointwise).  This is equivalent to the inequality $D^{2}\varphi \geq -\Lambda \text{Id}$ in the sense of distributions.  (One direction comes from the fact that $D^{2} (\varphi + \frac{\Lambda}{2} \|\cdot\|^{2}) \geq 0$ in the sense of distributions, making $\varphi + \frac{\Lambda}{2} \|\cdot\|^{2}$ convex and $-\frac{\Lambda}{2} \|\cdot\|^{2}$ is certainly smooth with second derivative actually equal to $-\Lambda \text{Id}$.  The other direction comes a direct computation and properties of convex functions.)

Either definition shows that if $\varphi$ is semi-convex, then the mollified family $\{\varphi^{\epsilon}\}_{\epsilon > 0}$ is also semi-convex.  In either case, we get the same "semi-convexity constant" $\Lambda$.

(Throughout $\varphi^{\epsilon}$ is given by $\varphi^{\epsilon}(x) = \int_{\mathbb{R}^{d}} \varphi(y) \eta^{\epsilon}(x - y) \, dy$ and $\eta^{\epsilon} \in C^{\infty}_{c}(\mathbb{R}^{d})$ satisfies $\eta \geq 0$, $\eta(x) = \eta(-x)$, and $\int_{\mathbb{R}^{d}} \eta(x) \, dx = 1$.  We set $\eta^{\epsilon}(\xi) = \epsilon^{-d} \eta(\epsilon^{-1} \xi)$.  Most likely, I will not use the symmetry.  I will assume, though, that $\eta(x) = 0$ if $\|x\| \geq 1$.)

Approach Using Definition 1:  Here we have $\varphi = \psi + \eta$ so $\varphi^{\epsilon} = \psi^{\epsilon} + \eta^{\epsilon}$.  Recall that $\psi$ is convex and $\eta$ is smooth.  Hence $\eta^{\epsilon}$ is smooth and we see that all we have to prove is that $\psi$ convex implies $\psi^{\epsilon}$ convex.  (Hence we only really needed to know that $\varphi$ convex implies $\varphi^{\epsilon}$ convex.)

In the next part of the answer, I will show mollification preserves convexity using the "distributional" approach.  Hence in this part I'll prove it "by hand."

$\psi$ is convex so, for each $x, x' \in \mathbb{R}^{d}$ and $\lambda \in [0,1]$, we have
\begin{align*}
\psi^{\epsilon}((1 - \lambda)x + \lambda x') &= \int_{\mathbb{R}^{d}} \psi((1 - \lambda)x + \lambda x' - y) \eta^{\epsilon}(y) \, dy \\ &\leq \int_{\mathbb{R}^{d}} [(1 - \lambda) \psi(x) + \lambda \psi(x')] \eta^{\epsilon}(y) \, dy \\
&= (1 - \lambda) \psi^{\epsilon}(x) + \lambda \psi^{\epsilon}(x').
\end{align*}
This proves $\psi^{\epsilon}$ is convex.

Similarly, it is not hard to show that $D^{2}\eta^{\epsilon} \geq -\Lambda \text{Id}$ holds.

Approach Using Definition 2: We want to show that if $f \in C^{\infty}_{c}(\mathbb{R}^{d})$ and $f \geq 0$, then
\begin{equation*}
\int_{\mathbb{R}^{d}} \varphi^{\epsilon}(x) D^{2}f(x) \, dx \geq -\Lambda \int_{\mathbb{R}^{d}} f(x) \, dx \cdot \text{Id}.
\end{equation*}

Notice that if $f \in C^{\infty}_{c}(\mathbb{R}^{d})$ and $f \geq 0$, then
\begin{align*}
\int_{\mathbb{R}^{d}} \varphi^{\epsilon}(x) D^{2}f(x) \, dx &= \int_{\mathbb{R}^{d}} \int_{\mathbb{R}^{d}} \varphi(x - y) D^{2} f(x) \, dx \, \eta^{\epsilon}(y) \, dy \\
&= \int_{\mathbb{R}^{d}} \int_{\mathbb{R}^{d}} \varphi(\xi) D^{2} f(\xi + y) \, d\xi \, \eta^{\epsilon}(y) \, dy.
\end{align*}
Since $f(\cdot + y) \in C^{\infty}_{c}(\mathbb{R}^{d})$ for each $y \in \{\eta^{\epsilon} > 0\}$, we have
\begin{equation*}
\int_{\mathbb{R}^{d}} \varphi(\xi) D^{2} f(\xi + y) \, d\xi \eta^{\epsilon}(y) \geq - \Lambda \int_{\mathbb{R}^{d}} f(\xi + y) \, d \xi \, \eta^{\epsilon}(y) \cdot \text{Id}.
\end{equation*}
Thus,
\begin{equation*}
\int_{\mathbb{R}^{d}} \varphi^{\epsilon}(x) D^{2}f(x) \, dx \geq -\Lambda \int_{\mathbb{R}^{d}} \int_{\mathbb{R}^{d}} f(\xi + y) \, d \xi \, \eta^{\epsilon}(y) \, dy \cdot \text{Id} = -\Lambda \int_{\mathbb{R}^{d}} f(x) \, dx \cdot \text{Id}.
\end{equation*}


问:
Fubini's Theorem and Stochastic (Ito) Integrals
Edit: in the comments, it is pointed out by @Mag that Ito Integral can - in fact - be defined pathwise, rather than an L2 limit. A bit of googling has led me to these Oxford university notes here for example. My question: would it be after all possible to apply Fubini's theorem to Ito Integral, defined pathwise?

Original question: I wonder whether Fubini's theorem is applicable to taking the expectation of Ito integrals. It is a well known fact that Ito integral posesses the "martingale property", i.e. for any deterministic or stochastic process $H(\omega,t)$, we have:

$$\mathbb{E}\left[\int_{h=0}^{h=t}H(\omega,h)dW_h\right]=0$$

From the above, it is obvious that for many types of integrands, Fubini's theorem "doesn't work", for example, we have that $\mathbb{E}[W_t^2]=t$, but:

$$\mathbb{E}\left[\int_{h=0}^{h=t}W_h^2dW_h\right]=0\neq \int_{h=0}^{h=t}\mathbb{E}\left[W_h^2\right]dW_h =\int_{h=0}^{h=t}hdW_h$$

Question: so what is it that makes Fubini's theorem not applicable above? The usual two conditions for being able to apply Fubini are:

(i) $\mathbb{E}[|H(\omega,t)|]<\infty$

(ii) $H(\omega,t)$ is measurable

In case of $H(\omega,t):=W_t^2$, both conditions are satisfied: (i) clearly, $\mathbb{E}[|W_t^2|]<\infty \forall t \in \mathbb{R}$ and (ii) $W_t$ is adapted and measurable (with respect to the filtration it generates and if the entire probability space is generated by $W_t$, then indeed $W_t$ is "measurable" with respect to $\left(\Omega,\mathbb{P},\mathcal{F}_t\right)$).

答:
I think the process $X_t:=\int_0^t H_s\mathbb dW_s$ is not a martingale in general. It is under very weak assumptions for $H$ a local martingale and if you have for example $\mathbb E(X_t^2)=\mathbb E(\int_0^t H_s^2\mathbb ds)<\infty$ for all $t\geq 0$ it is also a martingale. Another sufficent condition would be $\forall t\geq0: \mathbb E(\sup_{s\leq t}|X_s|)<\infty$.

The thing I mentioned in the comments to the other answer, that Itô integral is able to be defined pathwise, was just intended as an additional information, that the argument that it is defined as $L^2$-limit, is not a sufficent agrument against it (not saying anything about whether it actually possible or not). But being definied pathwise means still dependent on the path resp. $\omega$. Being definied pathwise or as $L^2$-limit is not the relevant part of your problem.

The main problem why there is definitly no Fubini-Theorem for Itô integrals combinied with integral with the measrue $P$ is, that $\mathbb E(\int_0^t W_s^2\mathbb dW_s)=\int_\Omega\int_0^t W_s^2\mathbb dW_s\mathbb dP$ is simply a real number, whereas $\int_0^t \mathbb E(W_s^2)\mathbb dW_s$ is stochastic process definitly depending on $\omega$.

But I remember an exercise regarding a Fubini-Theorem for Itô integrals with a Lebesgue integral, but it was a bit different to your application: Let $T>0$, $B\subset\mathbb R^n$ bounded and $$H:(\Omega,[0,T],B)\to\mathbb R: (\omega,t,x)\mapsto H(\omega,t,x)$$ a stochastic process not only depending on the time $t$ but also on $x$. Then under appropriate assumptions holds
$$\int_0^T\int_BH(\omega,s,x)\mathbb dx\mathbb dW_s=\int_B\int_0^TH(\omega,s,x)\mathbb dW_s\mathbb dx.
$$

If you are working with an Itô process $X$ there won't be much changes in the things above. For $\mathrm d X_t=b_t\mathrm dt +\sigma_t\mathrm dW_t$ you can simply replace the differentials as you are used and you are again in the situation above.

I hope this helps you a little bit for your problem.


问:
All of the dihedral groups are factor groups of the infinite dihedral group.

Show that $\operatorname{Aut}(\Bbb{Z}) \cong \{\pm 1 \}$ and write $\alpha : \mathbb Z_2 \rightarrow \operatorname{Aut}(\Bbb{Z})$ for the nontrivial homomorphism. The semidirect product $\Bbb{Z} \rtimes_{\alpha} \Bbb{Z_2}$ is called the infinite dihedral group. Show that the dihedral groups $D_{2n}$ are all factor groups of the infinite dihedral group.

We know that $\Bbb{Z}$ only has two generators, $1$ and $-1$. So there are only two possibilities: sending $1$ to $1$ and sending $1$ to $-1$. So $\operatorname{Aut}(\Bbb{Z}) \cong \{\pm 1\}$. There is an example in our textbook that shows that $\Bbb{Z}_n \rtimes_{\alpha} \Bbb{Z}_2 \cong D_{2n}$. So I only need to show that $$(\Bbb{Z} \rtimes_{\alpha} \Bbb{Z}_2)/\langle n \rangle \cong \Bbb{Z}_n \rtimes_{\alpha} \Bbb{Z}_2$$ right?

Suppose that $\Bbb{Z} = \langle b \rangle$ and $\Bbb{Z}_2 = \langle a \rangle$. We can see that $(\Bbb{Z} \rtimes_{\alpha} \Bbb{Z_2})/\langle n \rangle = \{b^ia^j \text{mod n} : b^i \in \Bbb{Z}, a^j \in \Bbb{Z}_2\} = \{b^ia^j : b^i \in \Bbb{Z}_n, a^j \in \Bbb{Z}_2\} = \Bbb{Z}_n \rtimes_{\alpha} \Bbb{Z_2}$.

Is my answer correct? I'm just curious because it seems too short.

Thank you in advance

答:
Your answer has a couple of issues as discussed in the comments:

You are using $n$ to denote the pair $(n,0)$ (recall that the elements of $\mathbb Z\rtimes \mathbb Z_2$ are defined as pairs).

More importantly, you are basically arguing that $\mathbb Z\rtimes \mathbb Z_2/\langle n\rangle$ and $\mathbb Z_n\rtimes \mathbb Z_2$ are equal as sets, which isn't really true or relevant; you want to show that they are isomorphic as groups.

You are correct however in your argument that $\mathbb Z$ has only two automorphisms, and your intuition that $\mathbb Z\rtimes \mathbb Z_2/\langle (n,0)\rangle \cong \mathbb Z_n\rtimes \mathbb Z_2$ is also right. To prove this, we consider the map $f:\mathbb Z\rtimes \mathbb Z_2\to \mathbb Z_n\rtimes \mathbb Z_2$ defined by $f((a,b))=([a]_n,b)$. Clearly this is a surjective map with kernel $\langle (n,0)\rangle$. We have
$$\begin{align}
f((a,b)(c,d)) &=f((a+(-1)^bc,b+d))\\
&=([a+(-1)^bc]_n,b+d)\\
&=([a]_n+(-1)^b[c]_n,b+d)\\
&=([a]_n,b)+([c]_n,d)\\
&=f((a,b))f((c,d))
\end{align}$$
thus $f$ is a homomorphism, so by the First Isomorphism Theorem we have $\mathbb Z\rtimes \mathbb Z_2/\langle (n,0)\rangle \cong \mathbb Z_n\rtimes \mathbb Z_2$ as desired.


问:
Finding infinitely many primes of the form $4k+3$ in a specific sequence
Question: Let $m$ be a natural number. Show that there exists infinitely many primes $p$ of the form $p=4k+3$ such that $p$ divides at least one integer in the sequence $2^nm+1$; $n \in \mathbb{N}$.

My attempt: If $ p \mid 2^nm+1$ for some $n$, then we have
$ m \equiv \frac{-1}{2^n} \pmod{p}$ for some $n$. This is equivalent to $ m \equiv -2^{n'} \pmod{p}$ for some $n'$.
Hence, if we can show the existence of infinitely many primes of the form $4k+3$ of the sequence $m+2^n$, then we will be done. But this didn't seem any easier. Also,I realized that if $2$ is a primitive root modulo some $p$ then that $p$ divides some term of the sequence irrespective of $m$. But overall I couldn't get anything useful.

答:
Using the $p$-adic order function, let

$$n_1 = \nu_2(m), \; m = 2^{n_1}m_1 \tag{1}\label{eq1A}$$

i.e., $m_1$ is an odd integer.  Consider only integers $n_2 \gt n_1$, with $n = n_2 - n_1$. Using this, along what you've already shown, the original problem is equivalent to proving there's infinitely many primes of the form $4k + 3$ which divide values of the sequence

$$2^{n_2} + m = 2^{n_1}(2^{n_2-n_1} + m_1) = 2^{n_1}(2^{n} + m_1) \tag{2}\label{eq2A}$$

This shows we just need to consider the sequence $2^n + m_1$. In modulo $4$, compare the value with $n = 1$, i.e., $2 + m_1$, to $2^n + m_1$ for $n \ge 2$. The congruence values are $1$ and then $3$, or $3$ and then $1$. Note this means the parity of the number of prime factors of the form $4k + 3$ changes, with it being even when the value is $1$ modulo $4$ and odd when the value is $3$ modulo $4$.

Let $a$ be the number (possibly $0$) of distinct prime factors of $2 + m_1$ of the form $4k + 3$, with these being $p_i$ for $1 \le i \le a$. For some positive integer $b$ and exponents $r_i \ge 1$, this gives

$$2 + m_1 = b\prod_{i=1}^{a}p_i^{r_i} \tag{3}\label{eq3A}$$

Next, let

$$d_1 = \prod_{i=1}^{a}p_i^{r_i + 1} \tag{4}\label{eq4A}$$

Initialize $j = 1$, and then use Euler's totient function and Euler's theorem to get

$$e_{j} = 2^{\varphi(d_{j}) + 1} + m_1 \equiv 2 + m_1 \pmod{d_{j}} \tag{5}\label{eq5A}$$

The total number of factors of $p_i$, for $1 \le i \le a$, in $e_j$ is the same as $2 + m_1$. However, since $\varphi(d_{j}) + 1 \ge 2$ then, as discussed earlier, the parity of the number of factors of the form $4k + 3$ of $e_j$ is different, so there must be at least one more distinct prime factor the form $4k + 3$. In particular, note none of them are equal to any $b_i$ for $1 \le i \le a$, as well as to any $q_i$ for $1 \le i \le j - 1$. Choose $q_j$ among any of these new additional prime factors, and let

$$d_{j+1} = d_{j}q_{j} \tag{6}\label{eq6A}$$

Increment $j$ and then repeat the above procedure starting at \eqref{eq5A}.

Note for each $j \ge 1$, there is a new, distinct prime factor $q_j$ of $e_j$ in \eqref{eq5A} which is of the form $4k + 3$. Thus, repeating the above procedure an infinite number of times shows there are infinitely many different prime factors of the form $4k + 3$ among the values in the sequence $2^{n} + m_1$ and, thus, also in $2^{n}m + 1$.


问:
Why is the vector space of a representation of the cyclic group $C_2$ a direct sum?
In Representation theory: Introduction Richard Borcherds gives the example of $\mathrm{G} = \mathbb{Z}/2\mathbb{Z} = \{1, g\}$. He says that a (complex) vector space $V$ is the direct sum $V = V^+ \oplus V^-$ because any vector can be written as:

$$
v = \frac{v + gv}{2} + \frac{v - gv}{2}.
$$

I would like to understand why $v_+ := \frac{1}{2}(v + gv)$ form a subspace $V^+$. This confuses me since $v_+$ depends on $v$ but the subspace $V^+$ cannot depend on $v$.

答:
Let $\phi : \mathbb{Z}/2\mathbb{Z} \to GL(V)$ be a representation. As $\phi(g)^2 = \phi(g^2) = \phi(1) = \operatorname{id}$, we see that $\phi(g)$ is diagonalisable with two possible eigenvalues, namely $1$ and $-1$. The splitting $V = V^+\oplus V^-$ is just the eigenspace decomposition where $V^{\pm}$ is the eigenspace of $\pm 1$. Explicitly, we can describe the subspaces by $V^{\pm} = \ker(\operatorname{id} \mp \phi(g))$; note that $v \in V^{\pm}$ if and only if $\phi(g)(v) = g\cdot v = \pm v$.

If $v \in V$, then $v = v^+ + v^-$ where $v^{\pm} \in V^{\pm}$. Then $g\cdot v = g\cdot v^+ + g\cdot v^- = v^+ - v^-$, so we see that $v^+ = \frac{1}{2}(v + g\cdot v)$ and $v^- = \frac{1}{2}(v-g\cdot v)$ which gives the decomposition of $v$ you provided.

An example you may have seen before is the case where $V$ is the vector space of all functions $\mathbb{R} \to \mathbb{R}$ and the $\mathbb{Z}/2\mathbb{Z}$ action is generated by $f \mapsto g\cdot f$ where $(g\cdot f)(x) = f(-x)$. In this case, $V^+$ consists of even functions and $V^-$ consists of odd functions. Any function $f$ can be written as the sum of an even function $f^+$ and an odd function $f^-$ where $f^{\pm}(x) = \frac{1}{2}(f(x) \pm f(-x))$.


问:
Double Integral of absolute value
How should I calculate the integral of $\int _0^1\:\:\int _0^1\:\left|x-y\right|dxdy$?

I understand that I need to split in two parts: $x<y$ and $x\ge y$

But I don't understand how to calculate it.

答:
a drawing can surely help you. when $x<y$ the integral is in the triangle above the line $x=y$ thus you have

$$\int_0^1\Bigg[\int_x^1(y-x)dy\Bigg]dx$$

when $x>y$ the integral is defined in the triangle under the line $x=y$ thus you have

$$\int_0^1\Bigg[\int_0^x(x-y)dy\Bigg]dx$$

Calculate the two integral and get the sum as a result

Title: Double Integral of absolute value


问:
Minimum number of zeroes of a function $f$ that makes $\int^b_a t^k f(t)dt = 0$ for all $k=0,1,...,n$

Question: Let $a < b, n \in N,$ and $f : [a, b] \to \mathbb R$ be a continuous function such that for all
$k = 0,1,...,n$ we have $\int^b_a t^k f(t)dt = 0$.
Determine the minimum number of elements of the set $\{ x \in (a, b):  f(x) = 0\}$.

I don't have any idea how to start. I only know that $f(t)$ must take both positive and negative values for the integral to be $0$, then there is at least one $x$ that makes $f(x)=0$ as the function is continuous.

I also tried to use the MVT for integrals to see that

$$\forall k, \exists c_k, f(c_k) \int^b_a t^kdt = 0$$

but I still couldn't get any extra information after evaluating the definite integral.

答:
Suppose the claim holds up to some $ n $ for a particular continuous function $ f $. We may assume $ f $ only has finitely many zeroes in $ [a, b] $, say $ N $ zeroes, and we may also assume without loss of generality that $ f(a) \geq 0 $.

In this case, the set of zeroes of $ f $ in the interior form a discrete subset of $ (a, b) $, and we may separate the zeroes into two categories: zeroes where $ f $ has a different sign on two sides of the zero, and zeroes where $ f $ has the same sign. Enumerate the zeroes in the former category as $ \alpha_1, \alpha_2, \ldots, \alpha_i $ (of course $ i \leq N $), and consider the integral (multiplied by a factor $ -1 $ if $ f(a) = 0 $ and $ f $ is negative on a neighborhood of $ a $, but this doesn't affect the basic argument):

$$ \int_{a}^{b} f(x) \cdot \prod_{h=1}^i (\alpha_h - x) \, dx $$

It's easy to see that the integrand is $ \geq 0 $, is continuous, and only vanishes on finitely many points. These imply that the integral is $ > 0 $, and since we're integrating $ f $ against a polynomial of degree $ \leq N $, this proves that $ n < N $. In other words, if $ N $ is the number of zeroes of $ f $, then $ N $ must be at least $ n+1 $.

To see that this lower bound on the number of zeroes is tight, it's enough to notice that the orthogonal complement of $ \langle 1, t, t^2, \ldots, t^n \rangle $ in $ \langle 1, t, t^2, \ldots, t^{n+1} \rangle $ is spanned by some polynomial $ P $ of degree $ n+1 $. It clearly has at most $ n+1 $ zeroes in $ [a, b] $, and by the above argument it must have at least $ n+1 $, so we conclude that it has exactly $ n+1 $.


问:
energy function of surfaces
The arc length of a curve $g(t)$ on a Riemannian manifold $M$ is
\begin{align}
L(g)=\int_a^b <\dot{g},\dot{g}>^{\frac 1 2}dt
\end{align}
Its variational problem is equivalent to that of the energy function
\begin{align}
E(g)=\int_a^b <\dot{g},\dot{g}>dt
\end{align}

For surface on $M$, can we have a similar conclusion?

The variational problem of area is equivalent to that of the energy of the surface?

答:
I am not sure if there is a corresponding energy functional, but there does exist generalized length functional.

For any submanifold $N$ in $(M,g),$ we can define the area functional $A(N)$ as we know. This is a natural generalization of your $L,$ since the critical points of $L$ are geodesics, and the critical points of $A$ are minimal submanifolds.

By the way, when calculating the derivative of $A,$ we use $\sqrt{\det g}$ in its local expression, where $g$ is the Riemannian metric. Maybe we can take the square root out to get an energy (not sure if this quantity has been investigated).

Title: energy function of surfaces


问:
convergenge of Hausdorff dimension for a sequence of compact sets
Recall that

the Hausdorff distance (metric) is:

A metric in the space of subsets of a compact set $K$, defined as follows. Let $X,Y\subset K$ and let $D_{x,y}$ be the set of all numbers $\rho(x,Y)$ and $\rho(y,X)$ where $x\in X$, $y\in Y$ and $\rho$ is a metric in $K$. Then the Hausdorff metric $\operatorname{dist}(X,Y)$ is the least upper bound of the numbers in $D_{x,y}$.

the Hausdorff dimension is:
Let $(X,d)$ be a metric space. In what follows, for any subset $E\subset X$,
$\operatorname{diam}(E)$ will denote the diameter of $E$.

For any $E\subset X$, any $\delta \in ]0, \infty]$ and any $\alpha\in [0, \infty[$ we consider the outer measure
$$
\mathcal{H}^\alpha_\delta (E) :=
\inf \left\{ \sum_{i=1}^\infty (\operatorname{[diam}\, E_i)^\alpha :
E\subset \bigcup_i E_i \quad\text{and}\quad \operatorname{ diam}(E_i)< \delta\right\}.
$$
The map $\delta\mapsto \mathcal{H}^\alpha_\delta (E)$ is monotone nonincreasing and thus we can define the Hausdorff measure $\alpha$-dimensional measure of $E$ as

$$
\mathcal{H}^\alpha (E) := \lim_{\delta\downarrow 0} \mathcal{H}^\alpha_\delta (E)
$$
The map $\delta\mapsto \mathcal{H}^\alpha_\delta (E)$ is monotone nonincreasing and thus we can define the  Hausdorff α-dimensional
measure of E as
$$
\mathcal{H}^\alpha (E) := \lim_{\delta\downarrow 0} \mathcal{H}^\alpha_\delta (E).
$$

I wonder if the following is true:  If $A_n \to A$ in the Hausdorff distance, then $\operatorname{dim}_H A_n \to \operatorname{dim}_H A$, where $A$ is a compact subset of $[0,1]^2$.

答:
Unfortunately this is not correct. Let us define
$$A_n:=\{\frac{k}{n}|\ k=0,1,\ldots n\}\subset [0,1].$$
Then $A_n$ converges in the Hausdorff sense to $[0,1]$. Let us proof this claim:
Let $x\in[0,1)$. Then for all $n\in\mathbb{N}$ exists a $k\in\{0,1,\ldots,n\}$ such that
$$\frac{k}{n}\leq x < \frac{k+1}{n}.$$
Hence
$$\rho(x,A_n)\leq \min(|x-\frac{k}{n}|,|x-\frac{k+1}{n}|)\leq |\frac{k+1}{n}-\frac{k}{n}|=\frac{1}{n}.$$
Therefore
$$dist_{\mathcal{H}}(A_n,A)\leq \frac{1}{n}\rightarrow 0\mbox{ for }n\rightarrow\infty.$$

The Hausdorffdimension of the $A_n$ is zero, because $A_n$ consists of finitely many points, while the Hausdorffdimension of $[0,1]$ is 1, because $\mathcal{H}^1([0,1])=\mathcal{L}^1([0,1])=1$ (Here $\mathcal{L}^1$ is the one dimensional Lebesgue mesaure).


问:
Continous functions and supremum
this might be a very trivial questions that can be answered by pointing to a theorem or lemma, but I have trouble remembering something clearly. Give a continuous function $f(x)$ with $x\in \mathbb{R}^n$. Why is (or is):
$$
\sup_{x\in \mathbb{R}^n\backslash\{0\}}f(x)=\sup_{x\in \mathbb{R}^n}f(x)
$$
Some similar equation came up in a class about a continuous likelihood function (for a Gaussian distribution) where the supremum of the likelihood function over $\beta\in \mathbb{R}^n$ was equated to the supremum of the likelihood function with $\beta\in \mathbb{R}^n, K\beta\neq c$ where $K \in \mathbb{R}^{k\times n}$ is a contrast matrix and $c\in \mathbb{R}^k$.

答:
It's because $0 \in \mathbb{R}^n$ is a limit point of $\mathbb{R^n}\backslash 0$. To prove the claim, note that there is nothing to prove if the supremum of $f$ is not attained at $0$. Suppose then that $\sup_{x \in \mathbb{R}^n}f(x) = f(0)$. Continuity of the function implies that
$$ \lim_{x\to 0}f(x) = f(0).$$
As such, for all $\epsilon > 0$ there exists $x \neq 0$ such that $f(0) < f(x) + \epsilon$. This implies that
$$\sup_{x \in \mathbb{R}^n \backslash 0}f(x) > f(0)-\epsilon$$
for all $\epsilon$. Hence the two suprema are equal.

Title: Continous functions and supremum


答:
Hint: $$\left\{ (x,y) : 0\leq x\leq 1, 0 \leq y \leq 1\right\} = \left\{ (x, y) : 0\leq x \leq 1, 0 \leq y < x\right\} \cup  \left\{ (x, y) : 0\leq x \leq 1, x\leq y \leq 1\right\} $$
is disjoint, so integral equal to sum of integrals over these two sets.

Also, using geometric interpretation may help: this integral equal to doubled (because of symmetry) volume of pyramid with height 1 and right triangle with legs of length 1 in base.  So, integral equals $\frac{2}{3}\cdot\frac{1}{2}\cdot 1 = \frac{1}{3}$.

Title: Double Integral of absolute value

Body:
How should I calculate the integral of $\int _0^1\:\:\int _0^1\:\left|x-y\right|dxdy$?

I understand that I need to split in two parts: $x<y$ and $x\ge y$

But I don't understand how to calculate it.


问:
Geometric interpretation of ramification of prime ideals.
I am trying to understand geometrically the ramification of primes in a finite separable field extension.

Let $A$ be a Dedekind domain with fraction field $K$ and $L/K$ a finite separable field extension of degree $n$, and let $B$ be the integral closure of $A$ in $L$, which is then also a Dedekind domain. Let $X=\text{Spec}(B)$ and $Y=\text{Spec}(A)$, and consider the projection $X\to Y$ which sends a prime $\mathfrak{q}\in X$ over $\mathfrak{p}\in Y$ to $\mathfrak{p}$ (i.e., the map of affine schemes corresponding to the inclusion $A\to B$).

Let $\mathfrak{p}\in Y$ be a non-zero prime ideal in $A$. The fiber over $\mathfrak{p}$ is then the set of prime ideals $\mathfrak{q}_{i}\in X$ such that
$$ \mathfrak{p}B=\mathfrak{q}_{1}^{e_{1}}\cdots \mathfrak{q}_{g}^{e_{g}}$$
is the corresponding prime factorization in $B$ (uniquely determined, because $B$ is a Dedekind domain). Let $f_{i}=[\kappa(\mathfrak{q}_{i}):\kappa(\mathfrak{p})]$ be the corresponding residue degrees. Then:

$$\sum_{i=1}^{g}f_{i}e_{i}=n$$

We say that $\mathfrak{q}_{i}$ is ramified above $\mathfrak{p}$ if the ramification index $e_{i}$ is strictly greater than 1, and unramified if $e_{i}=1$. We say that $\mathfrak{p}$ is unramified in $B$ if all $e_{i}=1$; that $\mathfrak{p}$ splits in $B$ if all $e_{i}=f_{i}=1$; and we say that $\mathfrak{p}$ is inert in $B$ if $g=e_{1}=1$.

We can picture both $X$ and $Y$ as algebraic curves and $X$ as a ramified cover of $Y$. I am trying to understand the precise geometric meaning of all the previous definitions in terms of this geometric picture.

My professor said that ramified primes above $\mathfrak{p}$ are those points $\mathfrak{q}_{i}\in X$ where the fiber over $\mathfrak{p}$ meets $X$ in a tangent point (I understand that by fiber he means the vertical line above $\mathfrak{p}$ when we picture $Y$ as a straight line). I tried to match this interpretation with the example of $A=\mathbb{Z}$ and $B=\mathbb{Z}[i]$, based on the picture of $\text{Spec}(\mathbb{Z}[X])$ of Mumford:

Prime ideals in $\mathbb{Z}[i]=\mathbb{Z}[X]/(X^{2}+1)$ should correspond to prime ideals containing $(X^{2}+1)$. That is, they should correspond to points in the curve $V(X^{2}+1)$ drawn in the picture.

The picture above $(2)$ matches this interpretation perfectly, since $(2)$ ramifies in $\mathbb{Z}[i]$. And so does the picture above $(5)$, since $(5)$ splits into two primes in $\mathbb{Z}[i]$. But now I have a problem with the picture above $(3)$. There should be only one point, namely $(3,X^{2}+1)\in X$, but I guess Mumford just doesn't draw it. But even if he did, how would he draw this point without creating a singularity on the curve? Because, being the spectrum of a Dedekind domain, $X$ should be a regular curve, right? So how would we draw this point?

But if you draw this point in the naive way, introducing a singularity, suddenly the residue degree seems to make some geometric sense to me: it seems to be the number of parameters that we need to describe the curve around the corresponding point. How much of this naive impression is true? Is this just a coincidence? Because algebraically this doesn't make much sense to me. One should be able to describe the curve locally with a single parameter, because all the local rings are DVR. I will try to summarize all these doubts in the following question:

What is the precise geometric meaning of the ramification index and the residue degree?

Finally, if possible, I would like to hear some clarification comments on the source that started all these doubts (initially I was naively happy with my professor's interpretation): Neukirch's Algebraic Number Theory. In chpater I, Section 13 (One-dimensional Schemes) he draws this ramification situation:

I have several problems with this picture.

Shouldn't $X$ be a regular curve?

Why aren't the ramification points tangent, as my professor claimed them to be?

If this is really the picture, how do we distinguish geometrically between inert points and ramification points? For example, the point on the right. Is it inert or is it ramified?

Neukirch says that this picture is only a fair rendering of the algebraic situation when the residue fields in $A$ are algebraically closed, but this doesn't help much to answer the previous questions.

Thanks for the spending the time to read this long question. I am quite confused with this issue and I wanted to express my doubts as clear as possible.

答:
I believe the following holds: Given any finite extension $K:=\mathbb{Q} \subseteq L$ with rings of integers $\mathcal{O}_L$, we get a canonical map

R1. $\pi: S:=Spec(\mathcal{O}_L)\rightarrow T:=Spec(\mathbb{Z})$.

The extension $\mathbb{Q} \subseteq L$ is always ramified, hence there are primes $p \in \mathbb{Z}$ where the fiber $\pi^{-1}((p))$ is non-reduced. The map $A:=\mathbb{Z} \subseteq B:=\mathcal{O}_L$ is an integral extension of rings and $\mathcal{O}_L$ is a finite rank free $\mathbb{Z}$-module. Hence $A\subseteq B$ is a flat map of rings.
The relative module of Kahler differentials $\Omega^1_{B/A}$ is non-zero, hence the map $\pi$ is not etale. The support $Z\subseteq S$ of $\Omega^1_{B/A}$ is a closed subscheme, and since $\pi(Z):=W\subseteq T$ is a closed subscheme (the map $\pi$ is finite), we get an open set $U:=T-W$. I believe the induced map $\pi_U: \pi^{-1}(U) \rightarrow U$ is etale, since $\pi_U$ is flat and $\Omega^1_{\pi^{-1}_U(U)/U}=0.$ Here you need a relative version of Hartshorne Exercise III.10.3. Maybe the stacks project does this, but I do not have a precise reference.

The "geometric" picture seems to be: Over $U$ it follows the number of closed points in the fiber is constant and the fiber rings are reduced. Outside of $U$ this is no longer true.

You find a related question here:

Why isn't every finite locally free morphism etale?


问:
Angle between pair of tangents drawn from a point to a conic

The angle between the pair of tangents drawn to the ellipse 3x^2 + 2y^2 = 5 from the point (1,2) is?

I considered using homogenization for this problem, consider the shifted coordinates:

$$ x' = x-1$$

$$ y' =  y-2$$

In shifted coordinates, our conic becomes:

$$ 3(x'+1)^2 + 2 (y'+2)^2 =5 \tag{1}$$

The relation of slope is given as:

$$ 3 (x'+1) + 2(y'+2) \frac{dy'}{dx} = 0 \tag{2}$$

Suppose the line passing the two intersection point is given as $ Ax+By =1$, then homogenizing (1) with it,

$$ 3(x'+1)^2 + 2(y'+2)^2 =  5 (Ax'+By')^2 \tag{3}$$

This factorizes to the form:

$$(y'-m_1x ) ( y'-m_2 x) = 0 \tag{4}$$

Now, how do I solve for $ \{m_1,m_2 \}$ from (3) and (4)? I'm not sure to how introduce (1) and (2) into helping me solve them.

Note: I've seen this question before already, but I want to solve it using homogenization.

答:
If you have the line pair as $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 $ then the angle is given by $\tan \theta = {2 \sqrt{h^2-ab} \over a+b}.$

E.g. $9x^2-24xy-4y^2+30x+40y-55=0$ and $\tan \theta = {2 \sqrt{12^2+9\cdot 4} \over 9-4}=\frac{12}{\sqrt{5}}.$


问:
Which law of logical equivalence says $P\Leftrightarrow Q ≡ (P\lor Q) \Rightarrow(P\land Q)$
I'm going through the exercises in the book Discrete Mathematics with Applications. I'm asked to show that two circuits are equivalent by converting them to boolean expressions and using the laws in this table.

$$\begin{array}{lcc}
\hphantom{1}\mathsf{1.\; Commutative\; laws:}       &  p\land q \equiv q\land p                          &  p\lor q \equiv q\lor p\\
\hphantom{1}\mathsf{2.\; Associative\; laws:}       &  (p\land q)\land r \equiv p\land (q\land r)        &  (p\lor q)\lor r \equiv p\lor (q\lor r)\\
\hphantom{1}\mathsf{3.\; Distributive\; laws:}      &  p\land (q\lor r) \equiv (p\land q)\lor (p\land r) &  p\lor (q\land r) \equiv (p\lor q)\land (p\land r)\\
\hphantom{1}\mathsf{4.\; Identity\; laws:}          &  p\land t \equiv p                                 &  p\lor c \equiv p\\
\hphantom{1}\mathsf{5.\; Negation\; laws:}          &  p\lor \lnot p \equiv t                           &  p\land \lnot p  \equiv c\\
\hphantom{1}\mathsf{6.\; Double\; negative\; law:}  &  \lnot(\lnot p) \equiv p                           &\\
\hphantom{1}\mathsf{7.\; Idempotent\; laws:}        &  p\land p \equiv p                                 &  p\lor p \equiv p\\
\hphantom{1}\mathsf{8.\; Universal\; bound\; laws:} &  p\lor t \equiv t                                  &  p\land c\equiv c\\
\hphantom{1}\mathsf{9.\; De\; Morgan}\text{'}\mathsf{s\; laws:}     &  \lnot(p\land q) \equiv \lnot p\lor \lnot q        &  \lnot(p\lor q) \equiv \lnot p\land\lnot q\\
\mathsf{10.\; Absorption\; laws:}                   &  p\lor (p\land q) \equiv p                         &  p\land (p\lor q) \equiv p\\
\mathsf{11.\; Negations\; of\; t\; and\; c:}        &  \lnot t \equiv c                                  &  \lnot c \equiv t\\
\end{array}$$

so as which law/s of logical equivalence says $P\Leftrightarrow Q ≡ (P\lor Q) \Rightarrow(P\land Q)$

I can see their equivalence clearly with a truth table. But the book is asking me to show it using the equivalence laws in the above table, and I can't see how any of them apply here. So, do any of those laws apply here in a way I'm not understanding? Or is there some other known law that does apply here?

答:
Let's assume the following definitions:

$\begin{array}{lc}
\mathsf{12.\; Definition \; of \; Implication:}       &  p \Rightarrow q \equiv \neg p \lor q \\
\mathsf{13.\; Definition \; of \; Biconditional:}     &  p \Leftrightarrow q \equiv (p \Rightarrow q) \land (q \Rightarrow p)
\end{array}$

Then we have:
$$\begin{array}{rll}
P \Leftrightarrow Q &\equiv (P \Rightarrow Q) \land (Q \Rightarrow P) &\text{by (13)} \\
&\equiv  (\neg P \lor Q) \land (\neg Q \lor P) &\text{by (12)} \\
&\equiv  (\neg P \land (\neg Q \lor P)) \lor (Q \land (\neg Q \lor P)) &\text{by (3)} \\
&\equiv  ((\neg P \land \neg Q) \lor (\neg P \land P)) \lor ((Q \land \neg Q) \lor (Q \land P)) &\text{by (3)} \\
&\equiv  ((\neg P \land \neg Q) \lor (P \land \neg P)) \lor ((Q \land \neg Q) \lor (P \land Q)) &\text{by (1)} \\
&\equiv  ((\neg P \land \neg Q) \lor c) \lor (c \lor (P \land Q)) &\text{by (5)} \\
&\equiv  ((\neg P \land \neg Q) \lor c) \lor ((P \land Q) \lor c) &\text{by (1)} \\
&\equiv  (\neg P \land \neg Q) \lor (P \land Q) &\text{by (4)} \\
&\equiv  \neg (P \lor Q) \lor (P \land Q) &\text{by (9)} \\
&\equiv  (P \lor Q) \Rightarrow (P \land Q) &\text{by (12)} \\
\end{array}$$