问:
Four Colour Theorem - What was wrong with De Morgan's 'proof'
Every resource I can find mentions his 'proof' but then just goes on to other ideas.

Is it because he could not prove his theory?

Surely it is trivial to prove using graph theory notation. (Or don't I know enough about graph theory to see why it isn't?)

答:
In chapter 2, pg. 17 of Four Colors Suffice: How the Map Problem was Solved
by Robin J. Wilson:

In this passage De Morgan hits upon the fact that if a map contains
four regions, each adjoining the other three, then one of them must be
completely enclosed by the others. He believed, incorrectly, that this
idea lay at the heart of the problem, and it soon became an obsession of
his. Since he could not prove it, he proposed to assume its truth as an
axiom, which he defined as “a proposition which cannot be made dependent upon obviously more simple ones.”

Specifically in De Morgan's letter to William Whewell, Master of Trinity College in  Cambridge in December 1853, he wrote:

I soon made out the following—which was at first incredible—then certainly
true—then axiomatic—for I cannot make it depend on anything I see more
clearly.

If four non-interfering compartments have each common boundary line
with the other three—one at least of the four must be inclosed by the other
three—or by fewer . . .

EDIT: @WizardMath, see below:

On pg. 167 of De Morgan on Map Colouring and the Separation Axiom by N. L. Bigg in the Archive for History of Exact Sciences volume 28, pages165–170 (1983), the axiom in modern terms could be summarised as follows:

Axiom. If four regions are mutually adjacent, then one is inclosed.

Theorem. Any plane map requires at most four colours.

Proof. (By induction.) The result is true for any map with not more than four regions. Suppose is it true for maps with at most $n$ regions, and let $M$ be a map with $n+1$ regions.

(i) If $M$ has four mutually adjacent regions, one of them is surrounded by the others. Removing (of shriking) this region results in a map $M'$ with $n$ regions, which can be four-coloured by the induction hypothesis. When the lost region is reinstated it is adjacent to at most three colours, and so there is a colour available for it.

(ii) If $M$ does not have four mutually adjacent regions, four colours are never needed in a 'neighbourhood' and so the induction step proceeds as in (i). Thus the theorem holds for each $n$.

Hope this helps.


问:
The asymptotics of the following expression
I wonder the asymptotics of the expression $$\sum_{k=1}^{n} \frac{n!}{kn^k(n-k)!}$$ when $n$ tends to be infinity and don't know how to do it. I have tried Stirling approximation but I think it doens't work. I can figure out that it's in $ O(\log n)$ scale but can't figure out its relation with $ \Theta(\log n / \log\log n)$. Can anybody give a guide on how to get its asymptotics? I conjecture that it doesn't exceed $ \Theta(\log n / \log\log n)$. Thanks a lot for any help!

答:
Let us denote your sum by $S(n)$. Making Maxim's calculations in the comments more precise, we can confirm Raymond Manzoni's asymptotics (with a more precise error term). Simple manipulation of Maxim's integral gives
$$
S(n) \sim \frac{{n!}}{{n^n e^{ - n} \sqrt {2\pi n} }}\frac{1}{2}\sqrt {\frac{n}{\pi }} \left( {\int_0^{ + \infty } {e^{ - nt} t^{ - 1/2} ( - \log t)dt}  - \log 2\int_0^{ + \infty } {e^{ - nt} t^{ - 1/2} dt} } \right)
$$
up to an absolute error which is $\mathcal{O}(n^{-1})$ times the leading order. By Theorem 2 on page 70 of R. Wong's book Asymptotic Approximations of Integrals, we have
$$
\int_0^{ + \infty } {e^{ - nt} t^{ - 1/2} ( - \log t)dt}  = \sqrt {\frac{\pi }{n}} \log n\left( {1 + \frac{{2\log 2 + \gamma }}{{\log n}} + \mathcal{O}\!\left( {\frac{1}{{\log ^2 n}}} \right)} \right).
$$
Also
$$
 - \log 2\int_0^{ + \infty } {e^{ - nt} t^{ - 1/2} dt}  =  - \log 2\sqrt {\frac{\pi }{n}}
$$
and, by Stirling's formula,
$$
\frac{{n!}}{{n^n e^{ - n} \sqrt {2\pi n} }} = 1 + \mathcal{O}\!\left( {\frac{1}{n}} \right).
$$
Collecting all the results, we finally obtain
$$
S(n) = \frac{1}{2}\log n + \frac{{\log 2 + \gamma }}{2} + \mathcal{O}\!\left( {\frac{1}{{\log n}}} \right).
$$
Taking more terms in the asymptotic expansion coming from R. Wong's theorem, we can replace the error term by an asymptotic expansion in inverse powers of $\log n$.

ContentLicense: CC BY-SA 4.0


问:
Polynomial in $z$ not bounded on $\mathbb{C}^-$
This is part of a proof that shows that explicit Runge-Kutta-methods are not A-stable. It states the following:
Given we have a polynomial $R(z)\in P_s$ in $z$  of degree at most $s\geq p>0$, then $R(z)$ is not bound on the set
$$
\mathbb{C}^-:= \{z\in \mathbb{C}: Re(z)\leq0 \}
$$
This was stated without a proof or explanation and is the only part in the proof I am struggeling with, so I would appreciate your help.

答:
I'm assuming $s>0$, as $0$ order methods don't make sense as explained to me in the comments. Polynomials are entire (holomorphic on the complex plane). So by Liouville's theorem, which states that a bounded entire complex function must be constant, $R(z)$ your polynomial must be unbounded on $\mathbb{C}$. To show it is unbounded on $\mathbb{C}^-$ you can first work with a monic polynomial and note it is unbounded on $\mathbb{C}^-$ by symmetry. To see this, suppose $z^m$ it is unbounded on just the right half plane. Then $(re^{i\theta})^m = r^m (\cos(m\theta)+i\sin(m\theta))$ is unbounded on the right half plane. Then reflecting across the imaginary axis yields $r^m(\cos(m\pi-m\theta)+i\sin(m\pi-m\theta))$ is also unbounded.

Since a polynomial $R$ of degree $s$ is such that $\lim_{|z|\to\infty} \frac{z^{s-1}}{R(z)}=0$ then $R(z)$ is unbounded, since $z^{s-1}$ is unbounded by the previous paragraph.


问:
How to determine that a set of elements of a linear space of $2\times3$ matrices is a linearly independent set
Let $E_1,E_2,E_3,E_4 \in \cal{M}_{2 \times 3}(\Bbb{K})$ be defined as follows.

$$E_1= \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix}, E_2= \begin{bmatrix}
    0 & 1 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix},$$ $$E_3= \begin{bmatrix}
    0 & 0 & 1 \\
    0 & 0 & 0 \\
\end{bmatrix}, E_4= \begin{bmatrix}
    0 & 0 & 0 \\
    1 & 0 & 0 \\
\end{bmatrix}.$$

$a)$ prove that set $\{E_i\}_{i \in I}$ is a linearly independent set,

$b)$ show that any element of $M$ can be represented as a linear combination of $\{E_i\}_{i \in I}$.

where $I = \{1,2,3,4\}$.

答:
Let $V$ be a vector space over a field $\Bbb{K}$ and $\{v_1,v_2,\dots,v_k\} \subseteq V$, where $k \in \Bbb{N}$. We say that the set $\{v_1,v_2,\dots,v_k\}$ is linearly independent, if the null vector $0_V$ can only be expressed as a linear combination of $v_1, v_2, \dots, v_k$ with all the coefficients null. So, we say that this set is linearly independent if
\begin{align*}
\alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_k v_k = 0_V \implies \alpha_1 = \alpha_2 = \dots = \alpha_k = 0.
\end{align*}

So, to answer the first question you have to start by writing the matrix $\cal{O}_{2 \times 3}$ as a linear combinations of $E_1, E_2, E_3, E_4$.
\begin{align}
\mathcal{O}_{2 \times 3} = \alpha_1 E_1 + \alpha_2 E_2 + \alpha_3 E_3 + \alpha_4 E_4.
\end{align}
This will give you (by solving the associated linear system), that $\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0$. Therefore $\{E_i\}_{i \in I}$ is a linearly independent set.

For the second question, I think that there is a mistake. In fact you can’t write every matrix of $\cal{M}_{2 \times 3}$ as a linear combination of the elements of $\{E_i\}_{i \in I}$. Note that $\dim{\cal{M}_{2 \times 3}(\Bbb{K})} = 6$. Therefore, $\langle E \rangle \ \neq \cal{M}_{2 \times 3}(\Bbb{K})$. For example, how would you write the matrix
\begin{align*}
A = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\end{align*}
as a linear combinations of elements in $\{E_i\}_{i \in I}$?.


答:
First of all, if statement $(b)$ is true, then it must be that $I = \{1,2,3,4,5,6\}$ where:

$$E_5= \begin{bmatrix}
    0 & 0 & 0 \\
    0 & 1 & 0 \\
\end{bmatrix}$$

$$E_6= \begin{bmatrix}
    0 & 0 & 0 \\
    0 & 0 & 1 \\
\end{bmatrix}$$

$(a):$ To determine if $\{E_i\}_{i \in I}$ is a linaerly independent set, we use the definition of linear independence. That is, $\{E_i\}_{i \in I}$ is a linearly independent set if and only if:

$$a_1E_1 + a_2E_2 + a_3E_3 + a_4E_4 +a_5E_5 +a_6E_6 = \begin{bmatrix}
    0 & 0 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix}$$

implies that $$a_1=a_2=a_3=a_4=a_5=a_6=0$$.

Thus, suppose that
$$A= \begin{bmatrix}
    a_1 & a_2 & a_3 \\
    a_4 & a_5 & a_6 \\
\end{bmatrix}= a_1E_1 + a_2E_2 + a_3E_3 + a_4E_4 +a_5E_5 +a_6E_6 =  \begin{bmatrix}
    0 & 0 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix}$$

It is then clear that $a_1=a_2=a_3=a_4=a_5=a_6=0$ and thus $\{E_i\}_{i \in I}$ is a linearly independent set.

(b) Let A be a matrix in $\cal{M}_{2 \times 3}(\Bbb{K})$. Then we have that:

$$A= \begin{bmatrix}
    a_1 & a_2 & a_3 \\
    a_4 & a_5 & a_6 \\
\end{bmatrix}$$

For some $a_i \in (\Bbb{K})$.

Now, we define a linear combination of $\{E_i\}_{i \in I}$ so that the coefficient of $E_i$ in the linear combination is the entry $a_i$ of the matrix $A$:

$$a_1E_1 + a_2E_2 + a_3E_3 + a_4E_4 +a_5E_5 +a_6E_6 = \begin{bmatrix}
    a_1 & a_2 & a_3 \\
    a_4 & a_5 & a_6 \\
\end{bmatrix} = A$$

Since $A$ was an arbitrary $2 \times 3$ matrix, $(b)$ holds.

Title: How to determine that a set of elements of a linear space of $2\times3$ matrices is a linearly independent set

Body:
Let $E_1,E_2,E_3,E_4 \in \cal{M}_{2 \times 3}(\Bbb{K})$ be defined as follows.

$$E_1= \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix}, E_2= \begin{bmatrix}
    0 & 1 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix},$$ $$E_3= \begin{bmatrix}
    0 & 0 & 1 \\
    0 & 0 & 0 \\
\end{bmatrix}, E_4= \begin{bmatrix}
    0 & 0 & 0 \\
    1 & 0 & 0 \\
\end{bmatrix}.$$

$a)$ prove that set $\{E_i\}_{i \in I}$ is a linearly independent set,

$b)$ show that any element of $M$ can be represented as a linear combination of $\{E_i\}_{i \in I}$.

where $I = \{1,2,3,4\}$.


问:
Lower semicontinuous of Minkowski functional
Let $P(x)$ is Minkowski functional of $C$ which is closed convex set and $\theta \in C$. After proving
$$
C=\{ x: P(x)\le 1\}
$$
and for any $\alpha >0$, $\alpha C$ is closed, how to get $P$ is lower semicontinuous  ?

I think the lower semicontinuous is
$$
\liminf\limits_{x\rightarrow x_0} P(x)=P(x_0).
$$
I can't see any connection between the lower semicontinuous and closure of $\alpha C$.

答:
Hints:

(1).  Show that for any $λ>0$, the set $\{ x \in X \colon P(x) \leq λ\}=λC $.

(2). Conclude that the sublevel sets $ L_λ= \{ x \in X \colon P(x) \leq λ\} $ are closed, for every $λ>0$. This is equivalent to saying that $P$ is lower semicontinuous.


答: But... $x^s$ is unbounded, isn't it ?

Title: Polynomial in $z$ not bounded on $\mathbb{C}^-$

Body:
This is part of a proof that shows that explicit Runge-Kutta-methods are not A-stable. It states the following:
Given we have a polynomial $R(z)\in P_s$ in $z$  of degree at most $s\geq p>0$, then $R(z)$ is not bound on the set
$$
\mathbb{C}^-:= \{z\in \mathbb{C}: Re(z)\leq0 \}
$$
This was stated without a proof or explanation and is the only part in the proof I am struggeling with, so I would appreciate your help.

Body: But... $x^s$ is unbounded, isn't it ?


问:
Discretising an inclined line and checking which range a particular point lies in.
I have the starting and end coordinates and equation of an inclined line. I want to discretize it into elements of length 1mm.
Then check in which range of elements a particular point lies.

The equation is in point-normal form: 6.16x - 9.23y -9.31 = 0.
The x and y coefficients are the components of the normal vector to the line.

My solution was:

To find the angle the line makes with the horizontal, using the formula for dot product since we know the components of the normal vector to the line.

Calculate the step size for the x and y coordinates considering the element size to be 1mm.
like this.

Take the point I want to check for and see which range it lies in.

Is there a better way to do this?

答:
Yes.

Let $P_0$ be your given "starting point".

Find the "direction vector" of the line. Let $\mathbf{v}$ be the vector pointing from the starting point to the ending point (obtained by subtracting coordinates).

Convert $\mathbf{v}$ to a unit vector. First, compute the length $h = \sqrt{9.23^2  + 6.16^2}$, and let $$
\mathbf{u} = \frac{1}{h} \mathbf{v}.
$$

Assuming that all the units here are in mm to begin with, you can, starting from $P_0$, construct the point
$$
P_i = P_0 + i \mathbf{u}.
$$
These points will lie along the line, separated by $1mm$. Use values $i$ from $0$ to $h$. if $h$ is not an integer, you will want to round up to the next higher integer.


问:
$\int |z| \bar z dz$ over contour
I was wondering whether I can receive some form of confirmation to my method as I feel it's very informal in the way it is worked out.

$$\int_{\Gamma+} |z| \bar z dz $$

Where $\Gamma +$ is a closed contour consisting of the upper semi-circle $|z|=1$,       $Im(z)\geq 0$ and the segment $-1\leq Re(z) \leq 1$, $Im(z)=0$ traversed in a positive direction.

The way I worked this out is essentially by splitting the closed contour into 2 contours and evaluating them seperately. My final result is that of $\pi i$.
Is my method correct? Or am I able to use some theorems such as Cauchy's Theorem or Green's Theorem? (Both of which I did not use as I am unaware of an easy way to show $f(z)=|z|\bar z$ is either holomorphic, or continuously differentiable)

答: Since $|z|=1$ on the integration path, you also have $|z|^2=z\overline z=1$ and thus $|z|\overline z=(1)(1/z)=1/z$.  So your differential is just $dz/z$ and the definite integral is $\ln [(-1)/(+1)]$.

ContentLicense: CC BY-SA 3.0

Title: $\int |z| \bar z dz$ over contour


问:
Sign of Cycle Notation
I know that the sign of cycle notation permutation of length $\displaystyle k$, will be: $\displaystyle ( -1)^{k-1}$.

The question is, what happens when in a cycle there are multiple orbits?

From Wikipedia: "The length of a cycle is the number of elements of its largest orbit.".

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

Some examples:

$\displaystyle ( 123)( 31) ,$ max $\displaystyle k=3$, $\displaystyle ( -1)^{3-1} =1$. On the other hand, I know this cycle has sign $\displaystyle -1$.

$\displaystyle ( 12)( 34)$, max $\displaystyle k=2$, $\displaystyle ( -1)^{2-1} =-1$. On the other hand, I know this cycle has sign $\displaystyle 1$.

Also, I know that each cycle can be written as an aggregiation of transformation (ie: $\displaystyle ( p_{0} p_{1} \dotsc p_{k}) =( p_{0} p_{1}) \cdotp ( p_{1} p_{2}) \cdotp \dotsc \cdotp ( p_{k-1} p_{k})$.

So under their assumption, I might argue each cycle has max lenght $\displaystyle k=2$, and therefor have sign $\displaystyle -1$.

A possible solution to this question, is that they referred to ONLY DISJOINT cycles. But here two questions arise:

In example 2 they were disjoint and still didn't work

And if so, then how can you quickly calculate the sign of not disjoint cycles? (without having to unroll them).

Bottom line, could anyway clarify how to calculate the sign of cycles with multiple orbits? Both for joints and disjoint ones.

Thanks

答: For example, consider that we dealing with $S_7$ and we have that $\sigma = \begin{pmatrix} 1 & 2 & 4 \end{pmatrix}$ and $\gamma = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 \end{pmatrix}$. The first is a cycle of length $3$. But $\gamma$ is not a cycle. So when using the property that the sign of a $k$-cycle is $(-1)^{k-1}$ you have to make sure that you have a cycle and not a product of cycles. For $\gamma$, the sign would be the product of the signs of each cycle. And you can prove that for all $\pi,\sigma \in S_n$ $\text{sgn}(\pi\sigma) = \text{sgn}(\pi) \text{sgn} (\gamma)$.

Title: Sign of Cycle Notation


问:
The optimization result change after shorten the equation
Assume that I have a function:

$$f_{w_1,w_2}(x)=w_2(w_1x)$$

I want to find the optimal point of function $f$, so I will follow the gradient:

$$\frac{\partial f}{\partial w_1} = w_2x, \frac{\partial f}{\partial w_2} = w_1x$$

Then I update the parameter:
$$\Rightarrow
\begin{cases}
w'_1 = w_1-w_2x \\ w'_2=w_2-w_1x
\end{cases}$$
$$\Rightarrow f_{w'_1w'_2}(x)=(w_2-w_1x)(w_1-w_2x)(x)$$

Now, let assume $w=w_1w_2$ then do the similar thing

$$f_w(x)=wx$$ $$\frac{\partial f}{\partial w}=x$$ $$w' = w-x$$ $$f_w(x) = (w-x)x$$

It's clearly that $f_{w'_1w'_2}(x)\ne f_w(x)$ which means by shortening the equation, we can get the better or worse optimization result. It sounds unclear to me, I think I am wrong somewhere, please tell me where to fix it.

答: If you set $w_1=w_2$ your equation should become $f_w=w^2x$


问:
Condition on $h^2-ab$ for deciding a conic

Consider the conic represented by the general equation $$ax^2+by^2+2hxy+2gx+2fy+c=0$$

It is known that if it's discrminant, that is, $abc+2fgh-af^2-bg^2-ch^2>0$, then it represents

1) A parabola, if $h^2=ab$

2) An ellipse, if $h^2<ab$

3) A hyperbola, if $h^2>ab$

I want to prove the above claim. My attempt:

Any two degree curve is either a pair of straight lines, a parabola , an ellipse (or a circle) or a hyperbola.

We start by considering the equation of conic as a quadratic in $x$. Solving for $x$ yields $$x=\dfrac{-(hy+g)}{a} \pm \dfrac{1}{a}\sqrt{(h^2-ab)y^2+2(gh-af)y+g^2-ac}$$

Now we want to find the conditions from above for it represents different conics. Suppose we want to find the condition of parabola. We know here $2$ things. There exists either a unique value of $x$ for a given $y$ (that is to say, it's vertex) or a unique value of $y$ for a given $x$, and secondly, parabola is an unbounded curve. So, what I am trying to do is to consider the graph of the quadratic $(h^2-ab)y^2+2(gh-af)y+g^2-ac$  and I want to prove somehow from here that $h^2=ab$ would result in a parabola along the lines I described, but till now I am unable to do so. Similarly, if it should be an ellipse, then since we know that ellipse is a bounded curve, somehow we need to prove that the graph of the quadratic equation $(h^2-ab)y^2+2(gh-af)y+g^2-ac=0$ would become a downward parabola with distinct real roots.

Thank you for your help.

答: Clearly, if $h^2-ab>0$, $y$ is unbounded and the curve is open (hyperbola). It negative, it is bounded and closed (ellipse). If zero, $y$ is unbounded only on one side.


问:
Self-intersection number of a complex curve in complex projective space
I'm currently trying to get a grip on actually calculating some differential-geometric definitions. I'm looking at the following map from $\mathbb{CP}^{1}$ to $\mathbb{CP}^2$ :

$f([z_0,z_1])=[z_0^3,z_0 z_1^2,z_1^3]$

What I don't understand is how one would go about calculating the self-intersection number of this. It's not an immersion, so do I need to take an immersion in the same homology class first? And how do I actually calculate it - by perturbing the map a little and then counting transversal intersections with sign (this turned out to be very messy with my choices), or by actually finding the Poincaré dual (how would one go about that?) and integrating it?

In case anyone is wondering, this problem arose while trying to understand the adjunction inequality for J-holomorphic curves.

Thanks already for any help you can give!

答: One has $g=(d-1)(d-2)/2$ for smooth curves so your curve is necessarily singular, as you point out.  But you don't need an immersion to calculate self intersection.  Since the homology class of the curve is $3[\mathbb{C}P^1]$, its selfintersection is necessarily $9$.  This is  because a projective line has self-intersection $1$ by definition of projective geometry!</p>


问:
consider a semicircle with a diameter AB where the length is 4, and a point C on the circular arc.
if the area of triangle ABC is  a half of the  maximum  and point C  is nearer to point A than point B  then the angle CAB  is? I have already found out that the  maximum area of triangle ABC is 4.But I have no idea on how to continue.</p>

答:
Let AB=c=2R=4, AC=b and BC=a, we have:

$S_{ABC}=\frac 12 (\frac 12 \pi R^2)=\pi$

Now we use this formula :

$$\frac{a+b}{c=4}=\frac{\cos \frac{A-B}2}{\sin \frac C2=\sin \frac{90}2=\frac{\sqrt 2}2}$$

$\cos(\frac{A-B}2)=\cos \frac A2\cos \frac B2+\sin \frac A2 \sin \frac B2$

$\cos(\frac{A+B}2)=\cos \frac A2\cos \frac B2-\sin \frac A2 \sin \frac B2$

Summing these relations we get:

$2\cos \frac A2\cos \frac B2=\cos(\frac{A-B}2)+\cos \frac {\pi}4$

Using relation:

$\frac a{\sin A}=\frac b{\sin B}=2R=4$

we finally get:

$\cos\frac A2\cos\frac B2=\frac{\sqrt 2}4(1+\sin A+\sin B)$

Considering $A+B=90^o$, we get $A\approx 75^o$.


问:
Expressing $ (\mathbb{Z}/2^k\mathbb{Z})^{\times} $ in terms of powers of $5$
I was doing my preview while some problems turned out to make me feel sucky.

It's said that every element $ x $ of $ (\mathbb{Z}/2^k\mathbb{Z})^{\times} $ could be expressed uniquely as
$$ x \equiv\pm 5^{\alpha} \bmod 2^k $$
where $ 1\le\alpha\le 2^{k-2} $, and the exponent $$ e=2^{k-2} $$
which I understood as the $ l.c.m $ of the orders of all $ x \in (\mathbb{Z}/2^k\mathbb{Z})^{\times} $ equals to $ 2^{k-2} $

All these above were written in my teacher's PPT for the number theory class. Since I'm doing a preview, I don't quite get that. Why is it and how to understand it?

Any explanation would be greatly appreciated!

答:
The following is an excerpt from a book whose draft I wrote last year. It answers your question in essentially the same way as in the comments by reuns, but in explicit detail.

Problem. Take the following steps for all integers $n\ge 3$:

Prove by induction on $n$ that $\nu_2 \left(5^{2^{n-2}}-1\right) =
    n.$

Show that the $\text{ord}_{2^n}(5) = 2^{n-2}.$

Deduce that
$S=\left\{\pm 5^k : k\in \left[2^{n-2}\right] \right\}$ is a reduced
residue system modulo $2^n.$

Solution. Each step leads to the next one.

We will show by induction on $n\ge 3$ that, for each integer $n\ge 3,$ there exists an odd integer $x_n$ such that $5^{2^{n-2}} = 1 + x_n \cdot 2^n.$ Since $x_n$ will be shown to be odd, no power of $2$ higher than $2^n$ can divide $5^{2^{n-2}}-1,$ while it is true that $2^n$ does divide $5^{2^{n-2}}-1.$ Thus, we will have proven that $\nu_2 \left(5^{2^{n-2}}\right) = n.$

The base case $n=3$ holds because $$5^{2^{3-2}} - 1 = 24 = 3\cdot 2^3.$$ Now suppose $5^{2^{n-2}} = 1 + x_n \cdot 2^n$ for some integer $n\ge 3$ and odd integer $x_n.$ Squaring the equation yields
$$5^{2^{n-1}} = 1 + x_n \cdot 2^{n+1} + x_n^2 \cdot 2^{2n} = 1 + x_n(1+x_n 2^{n-1})\cdot 2^{n+1}\equiv 1\pmod{2^{n+1}}.$$ Since $x_n$ is odd and $n\ge 3,$ so is $$x_{n+1}=x_n(1+x_n 2^{n-1}).$$

We know from the first part that $$5^{2^{n-2}} \equiv 1\pmod{2^n}.$$ So the order of $5$ modulo $2^n$ must divide $2^{n-2}.$ Suppose, for contradiction, that there exists an integer $i$ such that $3\le i\le n$ and $5^{2^{n-i}}\equiv 1\pmod{2^n}.$ Squaring this sufficiently many times (taking it to an exponent of $2^{i-3},$ to be precise) we get $5^{2^{n-3}} \equiv 1\pmod{2^n}.$ So there exists an integer $y_n$ such that $$5^{2^{n-3}} = 1 + y_n 2^n.$$ Squaring this yields $$5^{2^{n-2}} = 1 + y_n 2^{n+1} + y_n^2 2^{2n} = 1+(2y_n + y_n^2 2^n)2^n,$$ which implies that the $x_n$ from the last part is the even number $2y_n + y_n^2 2^n$. This is a contradiction.

Note that the order of $5$ modulo $2^n$ is $2^{n-2}=\frac{\varphi(2^n)}{2}$ and that the elements of $S$ are all in fact units. We could show that $S$ consists of precisely all of the units (that is, invertible elements, which are the odd integers in this case) by showing that none of the $5^i$ coincide with the $-5^j$ modulo $2^n.$ If $5^i\equiv -5^j\pmod {2^n}$ then cancelling the smaller of the two powers from both sides (or either, if they are equal) would yield a power of $5$ congruent to $-1$ modulo $2^n.$ It suffices to show that $$5^k \not\equiv -1\pmod{2^n}$$ for all $k.$ If the congruence held, then we could reduce it to $$5^k \equiv -1 \equiv 3\pmod{4}$$ using $4\mid 2^n.$ This is contradictory because $$5^k \equiv 1^k \equiv 1\pmod{4}.$$ Therefore, there are no overlaps between the $5^i$ and $-5^j.$ This produces $$2\cdot 2^{n-2}=2^{n-1}=\varphi(2^n)$$ distinct units modulo $2^n,$ which is the maximal number and so this is the set of all units modulo $2^n.$


问:
Why by my method i am counting more than the given in each case of more than 3 tiles
Problem where i was encountered was this :

A $7 \times 1$ board is completely covered by $m \times 1$ tiles without overlap.

Each tile may cover any number of consecutive squares, and each tile lies completely on the board.

Each tile is either red, blue, or green.

Let $N$ be the number of tilings of the $7 \times 1$ board in which all three colors are used at least once.

For example, a $1 \times 1$ red tile followed by a $2 \times 1$ green tile, a $1 \times 1$ green tile, a $2 \times 1$ blue tile, and a $1 \times 1$ green tile is a valid tiling.

Note that if the $2 \times 1$ blue tile is replaced by two $1 \times 1$ blue tiles, this results in a different tiling.

What i did was cases made for total number of tiles:

like for 3 tiles by doing $x_1$+ $x_2$+$x_3$= 7 , where $x_1$,$x_2$,$x_3$ represents lengths of each tile , which should some up to 7 and then multiply by 3! For 3 colours  from this method for cases more than 3 tiles ,

i seems like counting wrong always, like for example for 5 tiles i did first $x_1$+...$x_5$ = 7 this tells the number of tiles combinations possible to be 6C4 , and now the tiles can be of two types (1,1,3) ,(1,2,2) , where each represents the colour combinations from three colours .

Now for first type its like choosing 1 , then another 1 ,then 3 so its this:  5C1 * 4C1 * 3C3 and then 3! Mulitpilcation For three colours , similairy next type it would be 5C1 * 4C2 * 2C2 * 3! .

But ik i am making some couting mistake like i need to do some number division or multiplicatiom here which leads to correct answer for every case.
Can anyone tell whats wrong in this case of 5 tiles ?

答:
You've over-counted by a factor of $2$ at the end.  You shouldn't be multiplying by $3!$ but by $3$.  When you assign colors in the pattern $(1,1,3)$, for example, you only need to decide which color occurs $3$ times; after that there's no choice.  In the pattern $(2,2,1)$, you only need to decide which color occurs once.

Another way to do this part of the problem is inclusion-exclusion.  There are $3^5$ ways to color $5$ tiles with $3$ colors.  We must subtract the ways that only use $2$ colors, so we have $3^5-3\cdot2^5$.  Now what about a coloring with one color?  It's been counted once originally, and subtracted twice, since one of the two-colorings doesn't include it, so we need to add it back in.  This gives a final answer of $$3^5-2\cdot2^5+3=150$$
Note that this is the same as $$\binom51\binom41\cdot3+\binom51\binom42\cdot3=150$$


问:
Reduce the following set to obtain basis of real vector space $\mathbb{C}$
Reduce the set $\beta = \{2i-1,2i+3,4 \} $ to obtain basis of real vector space $\mathbb{C}$

What I know:

In order to be basis of a vector space, the set under consideration must be linearly independent and spans real vector space $\mathbb{C}$.

Clearly the set $\beta = \{2i-1,2i+3,4 \} $ is not linearly independent so we have to reduce it. But how?

Can someone please suggest anything. Thank you.

答:
Let $\Bbb{C}$ be the vector space over $\Bbb{R}$ and $\beta = \{2i-1,2i+3,4\}$. We want to obtain a basis of $\Bbb{C}$ from the set $\beta$. First, note that the canonical basis of $\Bbb{C}$ is $(1,i)$ (because $\Bbb{C}$ is over the field $\Bbb{R}$). So, $\dim{\Bbb{C}} = 2$, what means that any basis of this vector space must have exactly two elements.

Now, let’s look at the set $\beta$. It is straightforward to see that $\beta$ is linearly dependent because
\begin{align*}
2i + 3 = (2i-1) + 4.
\end{align*}

Hence, $\langle \beta \rangle = \langle \{2i-1,4\} \rangle$. To see if $\{2i-4,4\}$ is a basis of $\Bbb{C}$ all you have to do is to check if the set is linearly independent. I believe that you are able to check this by yourself.

After this, you will conclude that the set is linearly independent, so the $\dim{\langle \{2i-1,4\} \rangle} = 2$. But, since $\dim{\Bbb{C}} = 2$, we have that $\langle \{2i-1,4\} \rangle = \Bbb{C}$. Therefore, you get a basis $(2i-1,4)$ of $\Bbb{C}$ from $\beta$.


问:
$[M,\mathbb CP^\infty]=[M,\mathbb CP^2]$ where $M$ is a smooth closed orientable 3-manifold!
Prove the above result, where $[X,Y]$ means the set of all homotopy classes of maps from X to Y, two topological spaces. I have answered it below.</p>

答:
Here I claim that the correspondence $F:[M,\mathbb CP^2]\rightarrow[M,\mathbb CP^\infty]$ defined by sending $[f]$ to the homotopy class of $f$ as a map into $\mathbb CP^\infty$, is a bijection. If two maps are homotopic as maps into $\mathbb CP^2$, then so are they as maps into $\mathbb CP^\infty$ so the correspondence is well-defined.

It is clear that every map $f:M\rightarrow\mathbb CP^\infty$ can be homotoped to lie in the canonically included $\mathbb CP^2$ by the facts that every differential manifold has the homotopy type of a CW complex and that every map of CW complexes is homotopy to a cellular map, i.e., the k-skeleton maps into the k-skeleton for each k. So basically the map at time 1 in the homotopy has image in $\mathbb CP^1\subset\mathbb CP^2$. This shows that every homotopy class in $[M,\mathbb CP^\infty]$ is represented by a map into $\mathbb CP^2$, which shows that the correspondence $F$ is surjective.

For injectivity, I have to show that if $f,g:M\rightarrow\mathbb CP^2$ are homotopic as maps into $\mathbb CP^\infty$, then so are they as maps into $\mathbb CP^2$, or that a homotopy of such maps, as maps into $\mathbb CP^\infty$ descends to that as maps into $\mathbb CP^2$

Equivalently, I have to show that a map $F:M\times[0,1]\rightarrow\mathbb CP^\infty$ which sends $X=M\times\{0,1\}$ into $\mathbb CP^2$, is homotopic rel $X$ to a map $G$ with image in $\mathbb CP^2$.

$M\times [0,1]$ is a 4-d complex formed by attaching a 4-cell by mapping the boundary onto $M\times\{0,1\}\cup e_0\times[0,1]$ by the appropriate attaching map, where $e_0$ is a 0-cell. Now $M\times[0,1]\setminus X$ has the 1-cell $e_0\times(0,1)$ and the interior of the 4-cell attached. By the compression lemma 4.6 in Hatcher's Algebraic Topology book, it boils down to showing that $\pi_1(\mathbb CP^\infty,\mathbb CP^2)$ and $\pi_4(\mathbb CP^\infty,\mathbb CP^2)$ vanish.

By the homotopy exact sequence for a pair, $\pi_1(\mathbb CP^\infty,\mathbb CP^2)=0$ and $\pi_4(\mathbb CP^\infty,\mathbb CP^2)\cong\pi_3(\mathbb CP^2)$.

One then applies the homotopy exact sequence for $S^1\hookrightarrow S^5\rightarrow\mathbb CP^2$ and computes $\pi_3(\mathbb CP^2)=0$ $\square$


问:
Integral over elliptical region, what is going wrong with change of variables?
I would like to find
$$\iint_R \sqrt{x^2+25y^2} dA$$
where $R = \{(x, y)|x^2 + 25y^2 \le 1 \}$ using change of variables. I made the substitution
$$x = 5 r \cos(\theta)$$
$$y = r \sin (\theta)$$
The Jacobian determinant is $5r$ and the new region should be $S = \{(r, \theta)|0 \le r \le 1 \text{ and } 0 \le \theta \le 2 \pi \}.$ Thus, our integral should be equal to
$$\int_0^{2\pi} \int_0^1 r \cdot(5r) dr d\theta = \dfrac{10 \pi}{3}.$$

However, the correct answer is $2 \pi/15$. Can anyone please tell me what is going wrong? Thank you!

答:
One problem is in the range of $r$. After substituting $x = 5r\cos(\theta)$ and $y = r \sin(\theta)$ into $x^2 + 25 y^2 \leq 1$, we can get
$$ 25 r^2 \cos(\theta) + 25 r^2 \sin(\theta) = 25 r^2 \leq 1 $$.
Since $r\geq0$, in total we get $0 \leq r \leq \frac{1}{5}$.

Another one is that $\sqrt{x^2 + 25 y^2}$ should be $5r$, but you put $r$ instead.


问:
Fitch Style Proof $\forall x (\neg A(x) \rightarrow B(d)) \therefore \forall x A(x) \lor B(d)$.
I really need some help with the following proof. I want to show that
$\forall x (\neg A(x) \rightarrow B(d)) \therefore \forall x A(x) \lor B(d)$.

I tried to proof it with an indirect proof but always ended with something else. I dont know which subproof  I should choose to start with. I know that I need one, because I cant do muss with the implication in my premise.

Maybe I can assume $ \neg (\forall x A(x))$ for my indirect proof?

Indirect proof means this (proof per contradiction):

答:
Using material implication you can substitute $\lnot A(x) \to B(d)$ with $ \lnot\lnot A(x) \lor B(d)$ and then, using double negation elimination, with $ A(x) \lor B(d)$. You have then
$$ \forall x \big(A(x) \lor B(d) \big)$$
which, using distributivity of the universal quantifier, is equivalent to
$$ \big(\forall x A(x)\big) \lor B(d) $$


问:
Contour integral of $z^2 + \bar z$ along C: $|z-z_0| = R$
I am trying to solve the following contour integral:

$\oint_c (z^2 + \bar z )dz$             with C: $|z-z_0|=R$

My initial thoughts were just splitting the integral in two parts, the first one being over $z^2$, which then would be 0 due to Cauchy's integral theorem. The second part with $\bar z$ made me doubt, since Cauchy's theorem only works for holomorphic functions (and $\bar z$ does not seem holomorphic to me). Could anyone help me compute this integral? Thanks.

答:
You can make that $\overline z$ term into a function with only an isolated singularity and then use the Residue Theorem.  To wit:

$|z-z_0|=R$

$|z-z_0|^2=(z-z_0)(\overline z-\overline{z_0})=R^2$

$\color{blue}{\overline z=\overline{z_0}+\dfrac{R^2}{z-z_0}}$

Then from the Resudue Theorem, the counterclockwise contour integral of $\overline zdz$ around the circle will be $2\pi R^2i$ or $2i$ times the area enclosed by the contour.  That product actually applies for a clockwise integral  of  $\overline z dz$ around any simple closed curve.


问:
$f'(x)=f(x)$ and $f(0)=0$ implies that $f(x)=0$ formal proof
How can I prove that if a function is such that $f'(x)=f(x)$ and also $f(0)=0$ then $f(x)=0$ for every $x$. I have an idea but it's too long, I want to know if there is a simple way to do it. Thanks!
Obviously in a formal way.</p>

答: We can solve differential equation $f'(x)-f(x)=0$. It's linear differential equation with constant coefficient, and we know how to solve it. Setting $f(x)=e^{\lambda x}$ we obtain $\lambda e^{\lambda x} - e^{\lambda x}=0$, or $\lambda - 1 =0$, which implies $\lambda = 1$. So our solution is $f(x)=c e^x$. But, we have condition $f(0)=0$, so $c e^0=0 \implies c=0$, so our final solution is $f(x)=0$. </p>


问:
Homogenic Linear System Proof
I need help approaching this question,

any help will be much appreciated.

Assume that a linear system exists over $\Bbb{R}$ so that its solution group is:
$$S=\{(a,a^2,b)|a,b \in \Bbb{R}\}$$
Prove that system is homogenic

答:
No linear homogeneous system can have a set of solutions $S$ of this form.

Indeed the solution set of a homogeneous linear system is a vector space, in particular is stable by scalar multiplication.

But, here is a counterexample: taking $a=b=1$, $(1,1,1) \in S$ whereas its double  $2*(1,1,1)=(2,2,2) \notin S$ because the second coordinate isn't the square of the first one.

Title: Homogenic Linear System Proof


问:
From the set $\mathbb{Z}_8=\mathbb{Z}/8\mathbb{Z}$ select all elements for which there exists an inverse element with respect to multiplication
I have a misunderstanding with what it means
select all elements for which there exists an inverse element with respect to multiplication
does it mean that an element from $\{0,1,2,3,4,5,6,7\}$ and for example $4\times2=8$ therefore it's the identity element? but doesn't that also apply for $4\times0$ for exapmle? or do i have to consider the elements $\{0,1,2,3,4,5,6,7\}$ as just numbers and $a^{-1}=\frac{1}{a}$?

so to be more precise i have difficulties understanding the set of elements that has an inverse element with respect to multiplication from the set $\mathbb{Z}/8$

I would appriciate any help!
Thanks in advance!

答:
You're confusing the identity for addition, $0$, and the identity for multiplication, $1$.

Any numbers, that multiplied together, give the identity for ADDITION are called "zero divisors", and are precisely the non-invertible elements of your ring/algebra. For example this is the set of $0$-determinant matrices in a ring/algebra of square matrices; or the divisors of 8 (2, and 4)  in $\mathbb{Z}/8\mathbb{Z}$. This is the reason why only $\mathbb{Z}/p\mathbb{Z}$ where $p$ is prime are fields, since only in these do you have the fact that you have no non-unit divisors of $p$ (no zero divisors, and thus all elements are invertible).


问:
For which real $a$ does the series $\sum_{k=0}^{\infty}\frac{1}{(1+6k)^{ia}}-\frac{1}{(5+6k)^{ia}}$ vanish?
The question is to find when will $L(\chi, s)$ of $\chi=\left(\frac{-12}{n}\right)$ vanish. So  $\left(\frac{-12}{n}\right)$ is defined as the Jacobi symbol for odd $n$ ($\chi($even$)$=0). Using quadratic reciprocity I got $$\left(\frac{-12}{n}\right)=\left(\frac{-1}{n}\right)\left(\frac{3}{n}\right)\left(\frac{4}{n}\right)=\left(\frac{n}{3}\right)$$ so we have $$L(\chi,s)=\sum_{n=1\pmod 3, n\,\text{odd}}\frac{1}{n^s}-\sum_{n=2\pmod 3, n\,\text{odd}}\frac{1}{n^s}.$$ And we are only interested in when $s=ia$. How can I find where $L(\chi, s)$ vanish for those $s$?

答:
Your character is $\chi(n)=(\frac{n}3)1_{2\ \nmid\ n}$, the Dirichlet series is $L(\chi,s)=\sum_{n\ge 1}\chi(n)n^{-s}$.

It doesn't converge anywhere on $\Re(s)=0$. You need the analytic continuation, for a non-trivial Dirichlet character it amounts to $L(\chi,it)=\lim_{\sigma\to 0^+}L(\chi,\sigma+it)$.

$$L(\chi,0)=\lim_{s\to 0} L(\chi,s)=\lim_{s\to 0} s\int_1^\infty (\sum_{n\le x} \chi(n))x^{-s-1}dx$$

The mean value of $\sum_{n\le x} \chi(n)$ is $\frac23$.

$\sum_{n\le x} \chi(n)-\frac23$ is periodic zero-mean implies that
$$\int_1^\infty (\sum_{n\le x} \chi(n)-\frac23)x^{-s-1}dx$$ converges for $\Re(s) > -1$. Whence
$$L(\chi,0)=\lim_{s\to 0} s\int_1^\infty \frac23 x^{-s-1}dx= \frac23$$

Next, following the same argument as for $\zeta(s)$ we can show that $L((\frac{n}3),s)\zeta(s)$ has no zeros on $\Re(s)=1$, whence $L((\frac{n}3),s)$ has no zeros on $\Re(s)=1$. The functional equation implies that $L((\frac{n}3),s)$ has no zeros  on $\Re(s)=0$, and $$L(\chi,s)=(1+2^{-s})L((\frac{n}3),s)$$ implies that its zeros on $\Re(s)=0$ are at $\frac{(2 \Bbb{Z}+1)i\pi}{\log 2}$.


问:
two numbers chosen from 0 to 1, expected val of larger number
If I pick two numbers in the interval (0,1), what is the expected value of the larger number?

Please help. I know that the expected value of one number is 0.5 but I'm not sure how to find the expected vlaue of the larger number here.

答:
Assume that the value of the larger number is $x$. The probability of this is equal to the probability that the other number is less than $x$. This probability is obviously $x $. Therefore the expected value of the larger number is:
$$2\int_0^1 x\cdot x\, dx=\frac23,
$$
where the factor 2 counts the ways to choose the larger number.


问:
Reaching the gamma function from the Taylor series for $e^x$?
The gamma function is expressed as $n! =\Gamma(n+1)= \int_{0}^{\infty}x^ne^{-x}dx$

which reminds me of the Taylor series for $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$

Unmathematically, this makes the gamma function make sense to me: you're effectively doing a 'rearrangment' to get $n!$ However, if I am right about my intuition, I'm feeling particularly foolish about it, as I can't for the life of me properly explain what is precisely going on to get you from one to the other.

So, to you more experienced mathematicians out there, am I correct about my intuition, and if so, could I possibly trouble you to tell me exactly how you get from the Taylor series to the gamma function?

答:
It is true that on one hand,

$$\int_0^\infty\frac{x^n}{n!}e^{-x}dx=1$$ and on the other

$$\sum_{n=0}^\infty\frac{x^n}{n!}e^{-x}=1.$$

But the integral is on $x$, while the sum is on $n$, and I just see a coincidence here.

You can modify the argument of the exponential to ensure convergence and establish for $s>1$

$$\sum_{n=0}^\infty\int_0^\infty\frac{x^n}{n!}e^{-sx}dx=\sum_{n=0}^\infty s^{1-n}=\frac1{s-1}=\int_0^\infty e^{(1-s)x}dx=\int_0^\infty\sum_{n=0}^\infty\frac{x^n}{n!}e^{-sx}dx.$$


问:
If two topological spaces have the same topological properties, are they homeomorphic?
Topological properties are investigated because we can show that two spaces are not homeomorphic by finding one property that holds in one space but not the other. But what if no topological property can distinguish two topological spaces? So I ask:

If two topological spaces have the same topological properties, must they be homeomorphic?

Edit: I don't really have any particular class of topological properties in mind, because what I am thinking is really every single topological property, whenever it is well-defined for a topological space. I just didn't know that the class of topological properties is so large, that even "homeomorphic to a space $X$" is itself a topological property, making my question trivial.

答: Two topological spaces are either homeomorphic or not. So, this is a classification problem. But finding whether two spaces are homeomorphic is much more difficult than finding when they are not homeomorphic. As, homeomorphism leaves some characteristics of the space such as connectedness, compactness, number of holes etc unchanged (invariant), therefore, we look for these characteristics and if any of them is different then two topological spaces are not homeomorphic. If all these invariants  are same, then it does not guarantee that they are homeomorphic. Then what should we do? we should look for or define other topological invariants and check if they are different.Otherwise, we should find a homeomorphism between the two spaces. If not, then we should invent or wait for a new theory which indirectly solves the problem.</p>


答:
$\def\fitch#1#2{~~~~\begin{array}{|l}#1 \\\hline #2\end{array}}$

To derive a disjunction you must either derive one of its disjuncts, or use a reduction to absurdity.   This requires the later approach.

To derive a contradiction under the premise and assumption of $\lnot((\forall x~A(x))\lor B(d))$, first derive $\forall x~A(x)$ then use disjunction introduction.

$$\fitch{\forall x~(\lnot A(x)\to B(d))}{\fitch{\lnot((\forall x~A(x))\lor B(d))}{\fitch{\boxed a}{~~\vdots\\A(a)}\\\forall x~A(x)\hspace{20ex}\forall\textsf{I}\\(\forall x~A(x))\lor B(d)\hspace{10.5ex}{\lor}\textsf{I}\\\bot\hspace{26ex}\lnot\textsf{E}}\\(\forall x~A(x))\lor B(d)\hspace{14ex}\textsf{IP}}$$

So, now just derive $A(a)$ for an assumed arbitrary term $a$.  (Hint: second verse is much the same as the first).

Title: Fitch Style Proof $\forall x (\neg A(x) \rightarrow B(d)) \therefore \forall x A(x) \lor B(d)$.

Body:
I really need some help with the following proof. I want to show that
$\forall x (\neg A(x) \rightarrow B(d)) \therefore \forall x A(x) \lor B(d)$.

I tried to proof it with an indirect proof but always ended with something else. I dont know which subproof  I should choose to start with. I know that I need one, because I cant do muss with the implication in my premise.

Maybe I can assume $ \neg (\forall x A(x))$ for my indirect proof?

Indirect proof means this (proof per contradiction):


问:
Parametric equations finding a parametric equation such that two persons end up at point $P$ at the same time
Josh starts at the point $A=(0,0)$ and is heading towards the point $P=(12,3)$ and he uses 3 hours to reach this point. Tommy is located at $B=(0,9)$. One hour after Josh starts to walk towards $P$, then Tommy starts to head towards point $P$ as well. I need to find a parametrization for the path to Tommy such that he and Josh ends up at $P$ at the same time.

Attempt:
First we need to find where Josh is after one hour, I simply took $\mathbf{x}=(0,0)+t(12,3)$ where $t\in\mathbb{R}$. Since he uses three hours to reach $P$ then after one hour $(t=1/3)$ Josh is located at $\mathbf{x}=(0,0)+1/3(12,3)=(4,1)$. I'm not sure how to make sure that they end up at $P$ at the same time.

答:
Here is one parametrization,

The path of Josh to point $P$ is given by,

$r_1(t) = (0,0) + \frac{t}{3}(12,3), 0 \leq t \leq 3$

Simplifying, $r_1 = (4t, t), 0 \leq t \leq 3$

As Tommy starts one hour later and reaches point $P$ in two hours, path of Tommy to point $P$ will be,

$r_2(t) = (0,9) + \frac{t-1}{2}(12 - 0,3 - 9), 1 \leq t \leq 3$

Simplifying, $ \ r_2(t) = (6t-6, 12-3t), 1 \leq t \leq 3$


问:
Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$
Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting the least possible odd $n$ in the expression (because we can easily find such a $k$, that for a bigger $n$ the expression above is divisible by $2^k$, but for a smaller $n$ it isn't). So the least possible $n=3$ (for $n=1$ expression is $0$) and we have $4 \cdot 16 \cdot 100 \cdot 82 = 25 \cdot 41 \cdot 2^9$, so the $k$ we are looking for is $9$.
I am rather weak in number theory so I hope the get some feedback if my solution is correct.

Body: It's $\,(\color{#0a0}{n^4\!+\!1})(\color{#c00}{n^4\!-\!1})^2$ so $\,n\,$ odd $\,\Rightarrow\begin{align}&\ \,2\mid \color{#0a0}{n^4\!+\!1}\\ &2^4\mid \color{#c00}{n^4\!-\!1}\end{align}\,$ since $\,\begin{align} n&\equiv \pm1\ \ \pmod{\!4}\\\Rightarrow \color{#c00}{n^4}&\equiv \color{#c00}{(\pm1)^4}\!\!\!\!\pmod{\!4^2}\end{align}$


问:
Proof that if sets $A$ and $B$ are finited then $A\cup B$ are also finited
I started with: $\exists x_1,r_1 A \subset K(x_1,r_1) $ and $\exists x_2,r_2 B \subset K(x_2,r_2)$. Let $r=r_1 + r_2 + d(x_1,x_2)$. We want to show that $A\cup B \subset K(x_1,r)$. Let's check if $d(x,x_1) < r$:

$x\in A \to d(x,x_1) < r_1 < r$

$x\in B \to d(x,x_1) \le d(x,x_2) + d(x_2,x_1) \le r_2 + d(x_2,x_1)  \le r$

Is this proof correct?

答:
Yes, the proof is correct, BTW, such sets are called bounded not "finited" (a non-existent word).

The second line should read:

$$x\in B \to d(x,x_1) \le d(x,x_2) + d(x_2,x_1) < r_2 + d(x_2,x_1)  < r$$

with strict inequalities, of course.


问:
Definition of Improper integral
Definition 1: We say a function $f$ is integrable on $[0,\infty)$ if $f$ is integrable on $[0,b]$ for all $b<\infty$, and $\lim\limits_{b\rightarrow\infty }\int_0^b f(x)dx <\infty$. Then we call such number as $\int_0^\infty f(x)dx$. Similarly, we can define integrable function $(-\infty, 0]$. Then we say a function $f$ is integrable on $(-\infty, \infty)$ if $f$ is integrable on both $[0,\infty)$ and $(-\infty, 0]$ and $\int_{-\infty}^\infty f(x) dx = \int_{-\infty}^0 f(x) dx + \int_0^\infty f(x)dx$.

I am wondering if the above definition is equivalent as the following:

Definition 2: We say a function $f$ is integrable on $(-\infty, \infty)$ if $f$ is integrable on $[a,b]$ for all $-\infty<a<b<\infty$, and $\lim_{a\rightarrow -\infty} \lim_{b\rightarrow \infty} \int_a^b f(x) dx <\infty$, and we call such number as $\int_{-\infty}^\infty f(x) dx$.

Also, it seems that definition 2 also depends on the order of the limit. I want to confirm if that is the case.

答:
Your definitions are definitely not equivalent, and as you say there is a problem with definition 2 considering that that the order in which you take the limit can affect which functions are integrable according to that definition. Moreover, the rate at which $a$ and $b$ approach $\infty$ will also be important. The point of choosing definition 1 is that when we say that a limit converges, we mean that it converges on whatever fashion we choose to approach the limit. This is not captured in definition 2, because different integrals will converge depending on how you take the limit and you have no way of compensating for this.

The alternative, described briefly by DonAntonio, is often called the Cauchy Principal Value. The Cauchy Principal Value can be defined differently depending on whether one considers integrals that are improper in the sense that the integral limits go to $\infty$ or integrals of functions over singularities. The idea is however always the same, namely to the take limit symmetrically. In the case of integrals of the type $\int_{-\infty}^{\infty} f(x)\, dx$, we define it as follows

$$\lim_{r \to \infty} \int_{-r}^{r} f(x) \, dx$$

If this limit exists, we call it the Cauchy Principal Value and write

$$\mathbf{P} \int_{-\infty}^{\infty} f(x) \, dx = \lim_{r \to \infty} \int_{-r}^{r} f(x) \, dx.$$

This allows you to assign values to integrals that do not converge in the general sense because we choose one concrete way of taking the double limit in your definition 2, namely symmetrically. You can read more about the CPV on Wikipedia (https://en.wikipedia.org/wiki/Cauchy_principal_value).

Title: Definition of Improper integral


问:
Self-intersection number of a complex curve in complex projective space
I'm currently trying to get a grip on actually calculating some differential-geometric definitions. I'm looking at the following map from $\mathbb{CP}^{1}$ to $\mathbb{CP}^2$ :

$f([z_0,z_1])=[z_0^3,z_0 z_1^2,z_1^3]$

What I don't understand is how one would go about calculating the self-intersection number of this. It's not an immersion, so do I need to take an immersion in the same homology class first? And how do I actually calculate it - by perturbing the map a little and then counting transversal intersections with sign (this turned out to be very messy with my choices), or by actually finding the Poincaré dual (how would one go about that?) and integrating it?

In case anyone is wondering, this problem arose while trying to understand the adjunction inequality for J-holomorphic curves.

Thanks already for any help you can give!

答:
Well, the cohomology/intersection theory of $\mathbb{CP}^2$ is rather simple. Its generated by the class of the hyperplane. So every curve is $dH$ where $H$ is the class of the hyperplane and $d$ is an integer.

Furthermore, $\mathbb P^2$ has no curves of class $dH$ for $d<0$. You've described a cubic: so its class is $3H$, and self intersection is $9H^2$.

So $H^2 = pt$. Either by your favourite argument from topology for the cup product or by observing that two (generic) lines in $\mathbb P^2$ intersect at a point transversally.

In general, actually finding the Poincare dual to some class and integrating can be a bit of a pain. $\mathbb P^2$ is actually not too bad though. I've essentially described it above in very vague terms. There's only one real class to think about which is the class of the hyperplane.


问:
metric exercises, topology
consider R with the usual metric and $A = (1,2]$ find:

$d(0,A)$

$d(\frac{3}{2}, A)$

$d(1,A)$

$d(Q, A)$

I do not pretend that they do my homework, only that they explain one of these to me with the steps to follow to solve the others since I have found very little information on the subject please.

答:
$d(0,A)= 1$ as $1 \le x$ for all $x \in A$, and as $a_n=\frac{1}{n} + 1 \in A$ and $d(0, a_n)\to 1$ as $ n \to \infty$.

$\frac{3}{2}\in A$ so $d(\frac{3}{2}, A)=0$.

Using the same points $a_n$: $d(a_n,1)=\frac1n \to 0$ so $d(1,A)=0$ and the last one is unclear (what is $Q$?) but $d(\Bbb Q,A)=0$ as witnessed by $\frac{3}{2} \in \Bbb Q \cap A$.

Title: metric exercises, topology


问:
Doubt in real and complex vector space
Does saying "V is a vector space over reals" same as real vector space?
For eg:

$$ V={M}_{2}\left( {\mathbb{C}} \right)\\ M_2 \text{ is a } 2 \times 2 \text{ matrix and } \mathbb{C} \text{ is the field of complex numbers}\\
            W= {\{\begin{bmatrix} a \; b \\
                 c \; d\end{bmatrix} | \; a= d' \}}$$
Here my book states that W is not a subspace of complex vector space $${M}_{2}\left( {\mathbb{C}} \right)$$but it is a subspace of real vector space $${M}_{2}\left( {\mathbb{C}} \right)$$
Does real vector space here means that a,b,c,d are real numbers but the chosen field for scalar multiplication is complex?

I am very confused. (here $\mathbb{C}$ is referring to the field of complex numbers)

答:
"Real" or "complex" here refers to the field of the scalars for your vector space.

Take the "real vector space $\mathbb{C}^1$", it's just the complex numbers with complex addition as the group law, and scaling understood as multiplication of complex numbers by a real numbers, but complex multiplication is forbidden (cause we're not in an $\mathbb{R}$-algebra, only in an $\mathbb{R}$-vector space). So it's really isomorphic to $\mathbb{R}^2$ in this case.

Title: Doubt in real and complex vector space


问:
In definite integration, do I have to watch out for the behaviour of a function between the limits?

Calculate -
$$\int^{1}_{-1}\frac{e^\frac{-1}{x}dx}{x^2(1+e^\frac{-2}{x})}$$

This is my approach-

Substitute $t=e^\frac{-1}{x}$

$$\therefore \int^{\frac{1}{e}}_{e}\frac{dt}{1+t^2}$$

which gives us $\frac{\pi}{2}- 2\arctan{e}$, but the answer given is $\pi- 2\arctan{e}$. I'm guessing that my wrong answer has something to do with the exponential term.

答:
You can start from the substitution $\;y=-\dfrac1x.\;$ Then
$$I=\int\limits_{-\infty}^{-1}\dfrac{e^y\,\text dy}{e^{2y}+1}
+\int\limits_1^\infty\dfrac{e^y\,\text dy}{e^{2y}+1}
= \int\limits_1^\infty\dfrac{e^{-y}\,\text dy}{e^{-2y}+1}
+ \int\limits_1^\infty\dfrac{e^y\,\text dy}{e^{2y}+1}
= 2\int\limits_1^\infty\dfrac{e^y\,\text dy}{e^{2y}+1} $$
$$= 2\int\limits_e^\infty\dfrac{\text de^y}{1+e^{2y}}
=2\arctan e^y\bigg|_1^\infty =\pi-2\arctan e,$$
without doubts in the result.


问:
How to show $P−λI$ is invertible for every $λ≠0,1$
I'm trying to show that for the projection $P$, $P−λI$  is invertible for every $λ≠0,1$.

The eigenvalues of $P$ are $0$ and $1$ and to determine the spectrum of $P$, I am trying to show that $P−λI$  is invertible, for $λ≠0,1$. Can someone help me with this? I am not sure how to show this invertibility. Thanks!

答:
Let $\alpha=\frac1\lambda$ and consider instead $\alpha P-I$. Then we have
$$(\alpha P-I)(\beta P-I)=(\alpha\beta-\alpha-\beta)P+I$$
which equals to $I$ if it happens that $\alpha\beta=\alpha+\beta$, which occurs exactly when $\beta=\frac\alpha{\alpha-1}$, so since $\alpha\ne 1$, $\alpha P-I$ is invertible.


问:
Probability, dependent Events
I am trying to understand some of the tasks and would be happy if someone could check my solution and possibly point out any mistakes.
In advance, I am not asking for a solution, because I would like to work it out by myself.

The purpose of this exercise is to check whether events A and B are independent of each other.

Three dices are thrown.
The events are defined as follows:

A: All dice have the same number of dots.

B: The total sum of the dices is less than 5.

Therefore is

$A = \left \{ \left ( 1,1,1 \right ), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6) \right \}$

$B = \left \{ \left ( 1,1,1 \right ), (1,1,2), (1,2,1), (2,1,1) \right \}$

Some intermediate results:

$P\left ( A \right ) = \frac{6}{216} = \frac{3}{108}$

$P\left ( not A \right ) = 1 - P(A) = 1 - \frac{3}{108} = \frac{35}{36}$

$P(B) = \frac{4}{216} = \frac{1}{54}$

$P(A\cap B) = \frac{1}{216}$

$P(notA\cap B) = \frac{3}{216}$

A and B are independent if
$P(B|notA) = P(B|A)$

$
P(B|A) = \frac{P(A\cap B)}{P(A)} = \frac{\frac{1}{216}}{\frac{6}{216}} = \frac{1}{6}
$

$
P(B|notA) = \frac{P(notA\cap B)}{P(notA)} = \frac{\frac{3}{216}}{\frac{35}{36}} = \frac{1}{70}
$

Thus, the events A and B are dependent.

Thank you in advance for your time.

答:
Yes, it seems correct to me.

There can be some more straightforward ways to verify that, you could:

verify if $P(A \cap B) = P(A)P(B)$ that is the definition of independence and perhaps the best way

verify what you did, or, more generally, that $P(B|A) = P(A)$ or $P(A|B)=P(A)$ (you can also condition on $\bar{A}$ or $\bar{B}$) as independence means that the other event gives you no other information besides what you already have

In general, if two events are independent, then, also all the possible combinations of negations of the events are independent, therefore you can verify equivalently with just one of cases and that is enough.

To summarize, the following statements are equivalent (I indicate with $ A \perp \!\!\! \perp B$ the independence between A and B):

$A \perp \!\!\! \perp B$

$\bar{A} \perp \!\!\! \perp B$

$A \perp \!\!\! \perp \bar{B}$

$\bar{A} \perp \!\!\! \perp \bar{B}$

$P(A|B)=P(A)$

$P(B|A)=P(B)$

So, based on the data that you have, you can pick the one that better suits them and use it.

Title: Probability, dependent Events


问:
Split two integers of module operation $\pmod{m}$
Given the following module operation:

$$8x + 51\pmod{10}$$

Details:

$x \equiv y \pmod{m}$ is by definition equivalent to  $m|(x−y)$.

$x \equiv y \pmod{m}$ , $x,y\in \mathbb{Z}, m \in \mathbb{Z^+}$ iff $a \pmod {m} = b \pmod{m}$.

Question 1: Can we split it as follows based on module details above:

$$8x + 51\pmod{10} = 8$$

$$51\pmod{10} = 1$$

Then,

$$8x + 1\pmod{10} = 8$$

Question 2: Is it possible to subtract numbers from the module operation $1$ from both sides:

$$8x \pmod{10} = 7$$

Question 3: Suppose we have the quantity $8x \pmod{10} = 7$, then what is the meaning that the inverse of $8 \pmod{10}$ is $7 \pmod{10}$? Given the inverse, how we can multiply both sides by it as follows as this was not covered in basic module operations in

$$7\cdot 8x \pmod{10} = 7 \cdot 7 \pmod{10}$$

We can conclude that

$$56x \pmod{10} = 7 $$

答:
Indeed,

$8x+51\mod 10 = [(8x\mod 10) + (51\mod 10)] \mod 10 = [(8x\mod 10)+1]\mod 10 = 8x+1\mod 10$.

That is, the modulo operation is compatible with addition (and multiplication).

The additional $\mod 10$ is necessary, e.g.

$18+16\mod 10 = [(18\mod 10) + (16\mod 10)] \mod 10 = 8+6 \mod 10 = 14\mod 10 = 4.$


问:
Permutation acting on subsets of the domain
Suppose a group $G \subseteq S_n$, where $S_n$ is the permutation group on $X = \{1, ... ,n\}$.
Consider two subsets $Y,Z \subseteq X$ with $|Y| = |Z| \leq n$.

Problem: Is there a permutation $g \in G$ that maps $g(Y) = Z$, i.e. $g(y) \in Z \; \forall y \in Y$ ?

I'm looking for an efficient method to check if there exists such a $g \in G$.

My current solution to this, is a simple iteration over all elements $g$ and checking if the permutation maps $Y \rightarrow Z$. I was wondering if there exists a more efficient method, which I could use (e.g. in GAP). My background is not in Group Theory, so I might be missing the correct term for this problem.

答:
A practical method for testing whether one set can be mapped to another by a permutation is a backtrack search. (Whether it is efficient depends on your definition -- it is not polynomial time and as far as I know no general polynomial time method is known.)
It uses the group structure (and conditions on where points must get mapped) to avoid testing all group elements. In practice, it performs quite well in most cases.

In GAP, you can find such elements  using RepresentativeAction, specifying an action OnSets (which in GAP must be sorted lists):

```<code>gap> G:=Group((1,2,3,4,5),(1,2,3));;
gap> RepresentativeAction(G,[1,3,4],[2,4,5],OnSets);
(1,2)(3,5)
</code>```

Body:
A practical method for testing whether one set can be mapped to another by a permutation is a backtrack search. (Whether it is efficient depends on your definition -- it is not polynomial time and as far as I know no general polynomial time method is known.)
It uses the group structure (and conditions on where points must get mapped) to avoid testing all group elements. In practice, it performs quite well in most cases.

In GAP, you can find such elements  using RepresentativeAction, specifying an action OnSets (which in GAP must be sorted lists):

gap> G:=Group((1,2,3,4,5),(1,2,3));;
gap> RepresentativeAction(G,[1,3,4],[2,4,5],OnSets);
(1,2)(3,5)


问:
Find a particular subset of $\mathbb{R}$
What kind of subset S of $\mathbb{R}$ that has the properties

$S$ is uncountable

$S \cap (1,3)$ is countable

答: How about $$ S = (-1,0)\cup \{1-\frac{1}{n} : n\in \mathbb{N}\}?$$ I think that should work. Clearly $\sup S = 1$ (because $\frac{1}{n} \to 0$ as $n\to \infty$); $1$ is obviously not in $S$; it’s uncountable because $(-1,0)\subseteq S$; and finally $S\cap (0,1) = \{1-\frac{1}{n} : n\in \mathbb{N}\}\setminus \{0\}$ is countable because it’s a subset of $\mathbb{Q}$.


问:
What's the meaning of "at most" in logic?
I am having difficulty in finding meaning of the words "at most" in mathematics or more precisely in logic. It probably is to some extent a language and/or philosophy problem.

For example when we say "we can do ... with at most $x$ ... ". Does that mean that we can't do it with more than $x$ or does it mean that we can do it with less than or equal to x, but maybe also more?

答:
"At most" means that many or less.

I can fit at most 5 people in my car.

The auditorium seats at most 100 people.

The algorithm will converge in at most 1000 iterations.

The express checkout can be used with at most 10 items.

In all of these cases, the action can or will performed with X items or fewer. If you have more than X, the action is not possible (seating 150 people in the auditorium), or alternatively, the action will not occur with more than X (the algorithm converging in 1200 iterations).


问:
If every irreducible element in $D$ is prime, then $D$ has the unique factorization property.
Suppose every irreducible element in a domain $D$ is prime.

I'm trying to prove this implication:

In a integral domain $D$, if $a=c_1c_2...c_n$ and $a=d_1d_2...d_m$
  ($c_i,d_i$ irreducible), then $n=m$ and up to order $c_i$ and $d_i$
  are associates for every $i$.

My Solution

For each $i$, $c_i$ divides $d_1...d_m$, since $c_i$ is irreducible, hence prime, $c_i$ has to divide some $d_j$, $j\le m$, but $c_i$ and $d_i$ are irreducible, so $c_i$ and $d_j$ are associate.

Is my solution is correct so far?

how can I prove $n=m$?

Thanks in advance

答:
You're doing great, now cancel and continue.  (all integral domains are cancellative)

More precisely: prove your claim by induction on $\min(m,n)$.


问:
A question about the probability that the max of N dependent gaussian random variables with a deterministic covariance matrix exceed a constant gamma
The problem is given as follows,

I have a set of gaussian random variables $\{T_1,T_2,...T_N\}$, the mean value for each random variable is $\mu_n=k_n F$, where $k_n$ is an unknown but deterministic integer and $k_n \in [0,K]$.

$F$ is a function that I want to optimize and $F>0$.

The covarianve for each of them is $1$.

The covariance matrix $M$ of the $N$ gaussian random variables is unknown but deterministic.

There is a probability that is defined as
\begin{equation}
P=Pr\{\max\{T_1,T_2,...T_N\}>\gamma\}
\end{equation}

The question is "Can I say that $P$ can obtain its maximum when $F$ achieves its maximum?"

答:
The random vector variable $\mathbf{T} =(T_1,...,T_N)$ follows the multivariate normal distribution
$\mathcal{N}_N \left(\mathbf{k}F;\mathbf{M}  \right)$ where $\mathbf{k}$ is the vector $(k1,...,k_N)$.

We have: $\mathbf{T} = \mathbf{k}F+\mathbf{Z}$ with $\mathbf{Z}$ follows the distribution $\mathcal{N}_N \left(\mathbf{0}_N;\mathbf{M}  \right)$ ($\mathbf{0}_N$ is the vector $0$ of dimension $N$). Let's denote $f_{\mathbf{Z}}(\mathbf{z})$ the PDF of $\mathbf{Z}$.
\begin{align}
P(F) &= 1- P(\max\{T_1,...,T_N\} \le \gamma) \\
  &= 1- P(T_1\le \gamma,...,T_N\le \gamma) \\
  &= 1- P(Z_1\le \gamma - k_1F,...,Z_N\le \gamma-k_NF) \\
  &= 1- \int_{\{\mathbf{z}\in \mathcal{D}(F)\}} f_{\mathbf{Z}}(\mathbf{z})d\mathbf{z} \tag{1}\\
\end{align}

The integral is  calculated over the zone $\{\mathbf{z}\in \mathcal{D}(F)\}$ where $\mathcal{D}(F)$ is defined as
$$\mathcal{D}(F)= (-\infty,\gamma-k_1F ) \times \ldots \times (-\infty,\gamma-k_NF )$$

For 2 values $F_1$ and $F_2$ such that $F_1 > F_2$, we have $\gamma-k_nF_1 < \gamma-k_nF_2 $ for all $n =1,...,N$. So
$$(-\infty,\gamma-k_nF_1 ) \subset (-\infty,\gamma-k_nF_2 )$$
or
$$\mathcal{D}(F_1) \subset\mathcal{D}(F_2)  \tag{2}$$

From $(2)$, we deduce $\int_{\{\mathbf{z}\in \mathcal{D}(F_1)\}} f_{\mathbf{Z}}(\mathbf{z})d\mathbf{z} < \int_{\{\mathbf{z}\in \mathcal{D}(F_2)\}} f_{\mathbf{Z}}(\mathbf{z})d\mathbf{z}$.

Hence, we can deduce from $(1)$ that

$$ P(F_1) > P(F_2) $$

Conclusion: if $k_n \ge 0$, $P$ obtain its maximum when $F$ achieves its maximum.


答:
you can take S to be all the negative real numbers and add the rationals between 0 and 1(exclusive),
it's obvious that S is uncountable, 1 isn't in S, and supS=1.
S's intersection with (0,1) will be all the rationals between 0 and 1, which is a countable set.</p>

Title: Find a particular subset of $\mathbb{R}$

Body:
What kind of subset S of $\mathbb{R}$ that has the properties

$S$ is uncountable

$S \cap (1,3)$ is countable


答:
If $\displaystyle a=\frac1{\lambda(1-\lambda)}$ and $b=-1/\lambda$, then
$$
  (P-\lambda I)(aP+bI)=I.
  \tag 1
  $$

As I suggested in a comment below a deleted question by the OP,  this can be easily  deduced by solving equation (1) in the
unknowns $a$ and $b$.

Title: How to show $P−λI$ is invertible for every $λ≠0,1$

Body:
I'm trying to show that for the projection $P$, $P−λI$  is invertible for every $λ≠0,1$.

The eigenvalues of $P$ are $0$ and $1$ and to determine the spectrum of $P$, I am trying to show that $P−λI$  is invertible, for $λ≠0,1$. Can someone help me with this? I am not sure how to show this invertibility. Thanks!


问:
Is $f(x,y,z)=x^2y^2z^2\sin(\frac{1}{xyz})$ when $xyz\neq 0$ and $0$ else differentiable everywhere?

Is $$f(x,y,z)=\begin{cases}
x^2y^2z^2\sin\left(\frac{1}{xyz}\right), & xyz\neq 0\\
0, &\text{else}\end{cases}$$
differentiable everywhere?

Here is my proof.

If $x,y,z\neq0$ then $f$ is differentiable as a composition of differentiable functions.
if $x=0$ and $y=u,z=v$ then:
\begin{align*}
\lim_{(x,y,z)\to(0,u,v)} \left|\frac{x^2y^2z^2\sin(\frac{1}{xyz})}{\sqrt{x^2+(y-u)^2+(z-v)^2}}\right|&\leq \lim_{(x,y,z)\to(0,u,v)} |y^2z^2|\left|\frac{x^2}{\sqrt{x^2+(y-u)^2+(z-v)^2}}\right|\\[5pt]
&\leq u^2v^2 \lim_{(x,y,z)\to(0,u,v)} \left|\frac{x^2}{\sqrt{x^2}}\right|\\
&=0
\end{align*}

and so $f=0+o(|(x,y,z)-(0,u,v)|)$ so $Df=0$ and $f$ is differentiable.

In the same way we can show that if $x,y=0,z=u\neq0$ then $f$ is differentiable at $(0,0,u)$ with differential $0$.

Also at $(x,y,z)=(0,0,0)$ the function over $|(x,y,z)-(0,0,0)|$  is bounded by $C\cdot r^5$ and so its differentiable at $(0,0,0)$ ans all in all its differentiable everywhere.

Is my solution or am I missing something? I think I am correct, although my multivariable calculus is a bit rusty hence the post.

答:
Your proof is right in principle (I didn’t read too carefully) but it can be much simpler. This is the composition $f\circ g$ of the differentiable function
$$ f:\mathbb R\to \mathbb R, \quad f(r)=\begin{cases}r^2 \sin1/r&r\neq0\\0&r=0\end{cases}$$
With the differentiable function
$$ g:\mathbb R^3 \to \mathbb R,\quad g(x,y,z)=xyz$$
By the multivariate chain rule, $f\circ g$ is differentiable.


问:
How to use MVT to prove the following problem
Let $f(x)$ be continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b),|f'(x)|\leqslant1$. prove that for any $x_1,x_2\in[a,b]$ establish$$|f(x_1)-f(x_2)|\leqslant\frac{b-a}{2}.$$
I want to use MVT.But I don't know how did the $\frac{1}{2}$.

答:
Preliminary remarks: $f(a) = f(b)$ means that the function can be extended to a continuous function on $\Bbb R$ with period $b-a$. Then either or $x_1$ and $x_2$ can be shifted left or right by $b-a$ such that the new arguments satisfy $|x_1' - x_2'| \le (b-a)/2$, and $f(x_1) - f(x_2) = f(x_1') - f(x_2')$. The technical problem is that the extended function is not necessarily differentiable everywhere. However, that idea can be used for the following

Proof: Assume that $a \le x_1 \le x_2 \le b$.

If $x_2 - x_1 \le (b-a)/2$ then the mean-value theorem gives
$$
 |f(x_1) - f(x_2)| \le x_2-x_1 \le \frac 12 (b-a) \, .
$$

Otherwise $x_2 - x_1 > (b-a)/2$ and then we use the additional condition that $f(a) = f(b)$:
$$
|f(x_1) - f(x_2)| = |f(x_1) - f(a) + f(b) - f(x_2)|\\
\le |f(x_1) - f(a)| + |f(b) - f(x_2)|  \\\le (x_1-a) + (b-x_2) < \frac 1 2 (b-a) \, .
$$


问:
consider a semicircle with a diameter AB where the length is 4, and a point C on the circular arc.
if the area of triangle ABC is  a half of the  maximum  and point C  is nearer to point A than point B  then the angle CAB  is? I have already found out that the  maximum area of triangle ABC is 4.But I have no idea on how to continue.</p>

答:
Let $O$ be the center of the semi-circle. Then $O$ is the midpoint of the diameter $AB.$ The maximal area for the triangle $ABC$ occurs when $C \equiv C_{max}$ so that $C_{max}O$ is perpendicular to $AB$. Then $ABC_{max}$ is right-angle isosceles triangle so $\angle \, BAC_{max} = 45^{\circ}$.

The triangle $ABC$ has half the area of $ABC$ when the distance of $C$ from $AB$ is half the distance of $C_{max}$ to $AB$, the latter being the length of $C_{max}O$. If $M$ is the midpoint of $C_{max}O$, then the perpendicular from $C$ to $AB$ is equal and parallel to $MO$, so $CM$ is perpendicular to $C_{max}O$. Therefore $CM$ is the orthogonal bisector of $C_{max}O$ and therefore the triangle $CC_{max}O$ is isosceles with $CO = CC_{max}$. But $CO$ and $C_{max}O$ are radii of the semi-circle, so $CO = CC_{max} = C_{max}O$, i.e. triangle $CC_{max}O$ is equilateral and therefore $\angle \, COC_{max} = 60^{\circ}$.
Hence $\angle \, CAC_{max} = \frac{1}{2} \, \angle \, COC_{max} = 30^{\circ}$. Finally,
$$\angle \, CAB = \angle\, CAC_{max} + \angle \, BAC_{max} = 30^{\circ} + 45^{\circ} = 75^{\circ}$$


问:
Construction and properties of an Abstract Wiener Process on $C(\mathbb{R}^n,\mathbb{R}^n)$
I would like to consider a generalization of the classical Wiener process, also known as Brownian Motion.

While the Brownian Motion is a process $W \colon [0,\infty) \times \Omega → \mathbb{R}$, I would like a Wiener process defined on a multi-dimensional domain, that is:
$$f \colon \mathbb{R}^n \times \Omega → \mathbb{R} $$

I use the notation $f$, since I see it as a random function, so I would like $f(\cdot, \omega) \in C(\mathbb{R}^n, \mathbb{R})$.

I read here at section 4.2 that it is possible to define a Gaussian measure also on non-separable Banach spaces, but it is always concentrated on a separable subspace.

Now, the space $C(\mathbb{R}^n,\mathbb{R})$ is not even a Banach space, but it is a complete metric space with the topology of uniform convergence in compact sets. With this metric (which is not a norm), it should also be separable.

These are my questions:

Can one construct a Wiener Process on the space $C(\mathbb{R}^n,\mathbb{R})$? Does one use the topology of uniform convergence in compact sets?

I would like that for each $x \in \mathbb{R}^n$, $f(x)$ had a Gaussian distribution with mean 0 and variance $|x|$. The covariance between $f(x)$  and $f(y)$ could be e.g. $\prod_{i=1}^n \min(x_i,y_i)$.

What can we say about the (separable) support of this Wiener process?

What properties can we show about the realizations of this process? For instance, are they continuous a.s.? Are they non-differentiable a.s. like the 1-dim. Brownian Motion?

For the 1-dimensional Brownian Motion, it holds the following formula:
$$ \lim_{t → \infty} \frac{W_t}{t} = 0 \qquad \textrm{ a.s. .}$$
Can we say something about the asymptotic behaviour of our abstract Wiener process on $\mathbb{R}^n$, for instance that
$$ \lim_{|x| → \infty} \frac{f(x)}{|x|} = 0 \qquad \textrm{ a.s. ?}$$

Thank you for your help!

EDIT: I think that the Brownian Sheets may be the thing I am looking for. Does anyone know a good reference for them?

答:
With the indicated choice of covariance then it seems this is indeed the Brownian sheet.
I think a good introduction to this type of random object is the book "Multiparameter Processes: An Introduction to Random Fields" by Davar Khoshnevisan. For a freely accessible reference, see also his lecture notes http://www.math.utah.edu/~davar/UW/notes.pdf</p>


问:
Showing a Ring and an Ideal are equal
Q: Let R be a commutative ring with unity. Prove that if A is an ideal of R and A contains a unit, then A=R.

This is my attempt at an answer:
It suffices to show that all the elements in R are in A.

Let a be and element of A, and r be an element of R.
Since A is an ideal of R then all elements ar and ra must be in A.

We know that A has a unit say 1, then we know 1r and r1 for all r in R, are in A.

Thus all elements in R are in A, therefore A=R.

答:
It's: $I=R \Leftrightarrow \exists x\in I: \; x\in U(R)$

The way "=>" is common
The other way "<=" is alike yours: if $x\in I$ and $x\in U(R),\;$ because $I$=ideal, then $1_{R}=x\cdot x^{-1}\in I$. Then because $I \trianglelefteq R$=commutative, $\forall r\in R: \; \; r = 1_{R}\cdot r = r \cdot 1_{R} \in I$ => $I=R$. (but eventhough its alike your proof).

Title: Showing a Ring and an Ideal are equal


问:
how Riemann pass from the Euler product to the expression with exponential...
In his famous article Riemann about Zeta hypothesis start from the Euler's product and rewrite it as an integral. Can someone explain me how this passage is done, because, it's evident for the great Riemann but not for me …in other word how Riemann does to pass from the first to the second equation?</p>

答: Is this what you are looking for? At the link is a detailed proof following the approach outlined in Riemann's paper.</p>


问:
$S = \{A \in \mathcal{M}_{2x2} \colon A^2 = -I_2\}$, find elements in the set such that...
Consider a set $S = \{A \in \mathcal{M}_{2x2} \colon A^2 = -I_2\}$, where A is a matrix of complex numbers and $I_2$ is the identity matrix. What are all the elements of $D= \{\det(A) \colon A \in S\}$?

So, this question is saying the set is composed by all the complex matrices such that, when squared, are the symmetric of the identity matrix. And we are supposed to find the determinants where each of the matrices are in $S$. I'm a bit lost. How do I proceed?

Body:
Hint If $A^2=-I_2$ then
$$
\det(A^2)=\det(-I_2)
$$


问:
Ratio of random variables MGF? Please help.
If we know the moment generating functions (MGFs) of the random variables $X$ and $Y$ to be $M_X(s)$ and $M_Y(s)$, respectively. The MGF of the sum $X+Y$ will $M_X(s)\cdot M_Y(s)$.

So what is the MGF of the ratio distribution $\frac XY$?

答:
Unfortunately there is no simple formula for that.  That is why it took a while for the Student $t$-distribution to be worked out, for example, as it is the ratio of a $Z$ normal variable and the square root of a chi-square random variable over its degrees of freedom.

Essentially you have to slog through the integrals on a case by case basis, I believe.


答:
The word "with" is a bit ambiguous here. Suppose you said

We can nurdle any doohickey with at most $n$ widgets.

There could be (at least) two different, almost opposite, contexts.

Widgets are something a doohickey has, which make it harder to nurdle. We are saying that provided a doohickey doesn't have too many widgets, we will be able to nurdle it.

Widgets are used for nurdling things. We are saying we can always nurdle whatever doohickey we want, and we can guarantee not using too many widgets to do so.

In both cases we are not claiming the converse. It is possible in case 1 that we can also nurdle some doohickeys which have more than $n$ widgets, but it is also possible that we can't (or that we just don't know). In case 2, we can always avoid using more than $n$ widgets, if we want to, but perhaps we could choose to be wasteful and use more than $n$.

(You can make the statement unambiguous by replacing "with" with "having" for the first meaning, or "using" for the second.)

Title: What's the meaning of "at most" in logic?

Body:
I am having difficulty in finding meaning of the words "at most" in mathematics or more precisely in logic. It probably is to some extent a language and/or philosophy problem.

For example when we say "we can do ... with at most $x$ ... ". Does that mean that we can't do it with more than $x$ or does it mean that we can do it with less than or equal to x, but maybe also more?


问:
Cartesian products involving empty sets
Let $(\Omega_i)_{i\in I}$ be a nonempty family of sets. For each $i \in I$, let $\mathcal{E}_i$ be a set of subsets of $\Omega_i$. Let

$$ \coprod_{i\in I} \mathcal{E}_i:=\bigg\{\prod_{i\in I}A_i : A_i\in\mathcal{E}_i \cup\{\Omega_i\} ,
 A_i\neq \Omega_i \text{ for finitely many }i\in I \bigg\}$$

Let $(I)_{\lambda \in \Lambda}$ be a partition of $I$ into disjoint nonempty sets. I am trying to prove that

$$\coprod_{i\in I} \mathcal{E}_i= \coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$$

when identifying sets of the form $\prod_{i\in I} A_i$ $(A_i\subset\Omega_i)$ with $\prod_{\lambda \in \Lambda} \big( \prod_{i\in I_\lambda} A_i\big)$. I already showed $\coprod_{i\in I} \mathcal{E}_i \subset \coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$, but I am not sure the reverse inclusion holds if $\Omega_i =\emptyset$ for some $i$.

Is this reasoning correct? Thanks a lot for your help.

答:
$\Omega_i = \emptyset$ does not cause a problem, both sides will just be empty.

Your definition of $\coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$ is incomplete, as you haven't defined an $\Omega_{I_\lambda}$ for each member of $\coprod_{i\in I_\lambda} \mathcal{E}_i$ to be a subset of. If we put that to be $\bigcup \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$ then your equality does hold.

To show the reverse inequality note that any member $\prod_{\lambda \in \Lambda} B_\lambda$ of $\coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$ will have, for all but finitely many $\lambda$, $B_\lambda = \Omega_{I_\lambda} = \prod_{i\in I_\lambda} \Omega_i$. So each member is of the form $\prod_{\lambda \in \Lambda} \big( \prod_{i\in I_\lambda} A_i\big)$ with $A_i \neq \Omega_i$ for only finitely many $i$  from only finitely many $I_\lambda$, and so it identifies with $\prod_{i\in I} A_i$ for only finitely many $A_i \neq \Omega_i$ overall.

Title: Cartesian products involving empty sets


问:
Is there a homomorphic mapping between natural language and logic
To me, this might be possible with a very complicated logic but then there should be no calculi that are useful. And is there an algorithm for it that could do that? This already looks very undecidable.</p>

答: I believe that does not exist, because natural language is inconsistent, has many definitions for the same words, connotations as well as denotations that vary with the speaker and the listener, and in general is nothing like logic.  Any language that has puns and poetry is not going to be describable with logic.</p>


问:
Every function is the sum of an even function and an odd function in a unique way
It is known that every function $f(x)$ defined on the interval $(-a,a)$ can be represented as the sum of an even function and an odd function. However

How do you prove that this representation is unique?

Thanks for your help.

答:
If $f = g_1 + g_2 = h_1+h_2$ where $g_1,h_1$ are even and $g_2,h_2$ are odd then
$$
  g_1 - h_1 = g_2 - h_2 \tag{1}
$$
where the left-hand side of $(1)$ is even and the right-hand side is odd, hence both sides are just $0$. Indeed, it is easy to show just from the definitions that any function which is both even and odd must be a constant $0$.</p>


问:
Proving that Gauss map $N: S \rightarrow \mathbb{S}^{2}$ is surjective
I'm taking a course in elementary differential geometry, and there was the following problem.

Let $S \subset \mathbb{R}^{3}$ be a non-empty, compact, and connected surface. Assume that (by the "Jordan-Brouwer" theorem) we can talk about "inside and outside" components of $S$. Let $N(p)$ be the outward normal vector at $p \in S$. Prove that the Gauss map $N: S \rightarrow \mathbb{S}^{2}$ is surjective.

Solution: Fix a unit vector $v \in \mathbb{S}^{2},$ and consider the family of planes
$$
P_{c}=\left\{x \in \mathbb{R}^{3} \mid\langle x, v\rangle=c\right\}, \quad c \in \mathbb{R}
$$
Since $S$ is compact, the function $x \mapsto\langle x, v\rangle$ on $S$ achieves its maximum value at some point $p \in S$. Define $c_{0}=\langle p, v\rangle$. Then $S \cap P_{c}$ is empty for all $c>c_{0}$ whereas $S \cap P_{c_{0}}$ is not empty. We claim that $v=N(p)$. To see this, consider an arbitrary smooth curve $\alpha:(-\epsilon, \epsilon) \rightarrow S$, with $\alpha(0)=p$ and $\alpha^{\prime}(0)=w \in T_{p} S$. Since $f(t)=\langle\alpha(t), v\rangle$ is maximised at $t=0$, we must have $0=f^{\prime}(0)=\langle w, v\rangle$. Therefore, $v$ is orthogonal to any tangent vector to $S$ at $p$. In fact, $v$ must be the outward unit normal vector, since all of $S$ lies in the half-space $\langle x, v\rangle \leq\langle p, v\rangle$. As $v \in \mathbb{S}^{2}$ was arbitrary, the Gauss map $N: S \rightarrow \mathbb{S}^{2}$ must be surjective.

I don't understand the line "In fact, $v$ must be the outward unit normal vector, since all of $S$ lies in the half-space $\langle x, v\rangle \leq\langle p, v\rangle$." Why is this? I can't understand this intuitively, nor can I understand it formally (with an actual proof). Does anyone know why this is?

答:
The catch lies in this part of the problem:

Assume that (by the "Jordan-Brouwer" theorem) we can talk about "inside and outside" components of $S$.

Note that this wasn't used elsewhere in the presented reasoning.

For an (at least) intuitive explanation, note that $P_{c_0}$ divides the whole space into two half-spaces:
$$
P^+_{c_0}=\left\{x \in \mathbb{R}^{3} \mid\langle x, v\rangle > c\right\},
\quad
P^-_{c_0}=\left\{x \in \mathbb{R}^{3} \mid\langle x, v\rangle \le c\right\}.
$$
According to the proof, all $S$ is contained in $P^-_{c_0}$, since $c_0$ was chosen as the maximal value of $\langle x, v \rangle$. Hence, the whole half-space $P^+_{c_0}$ is on the outside of $S$.

On the other hand, the vector $v$ (think of it as anchored at $p \in P_{c_0}$) points towards $P^+_{c_0}$, e.g. in the sense that
$$
\langle p+tv, v \rangle = c_0 + t > c_0
\quad \text{for } t>0,
$$
so each point $p+tv$ (with positive $t$) lies in $P^+_{c_0}$. This should explain why we call it the outer normal.


问:
Maximize $\langle {\bf A} , {\bf X} \rangle$ subject to $\| {\bf X} \|_* \leq 1$

Given ${\bf A} \in \mathbb R^{m \times n}$, $$\begin{array}{ll} \text{maximize} & \langle {\bf A} , {\bf X} \rangle\\ \text{subject to} & \| {\bf X} \|_* \leq 1\end{array}$$ where $\| \cdot \|_*$ denotes the nuclear norm.

Though I know something about the spectral norm, I know almost nothing about the nuclear norm,  dual norms, convex analysis, etc.  Since I am utterly unqualified to answer this on my own, I post this question.

Related:

Show that the dual norm of the spectral norm is the nuclear norm

Maximize $\langle \mathrm A , \mathrm X \rangle$ subject to $\| \mathrm X \|_2 \leq 1$

答:
This is easy to formulate using CVX, wbich will convert it to an SDP. How reliably and quickly it solves depends on the size and numerics of the problem.</p>
```<code>cvx_begin
variable X(m,n)
minimize(trace(A'*X))
norm_nuc(A) <= 1
cvx_end
</code>```

Body:
This is easy to formulate using CVX, wbich will convert it to an SDP. How reliably and quickly it solves depends on the size and numerics of the problem.

cvx_begin
variable X(m,n)
minimize(trace(A'*X))
norm_nuc(A) <= 1
cvx_end


问:
Determining convergence of $\frac{(x^2)}{x!}$
I'm trying to determine the convergence of the infinite series

$$\frac{x^2}{x!}$$

The factorial clues me in to using the ratio test, but I get stuck when trying to cancel.

答: That quotient is equal to $$\frac1{(3x+3)(3x+2)(3x+1)}\left(\frac{x+1}x\right)^6.$$Since$$\lim_{x\to\infty}\frac1{(3x+3)(3x+2)(3x+1)}\left(\frac{x+1}x\right)^6=0<1$$your series converges.


答:
Richard Montague tried something along these lines in the 1970s. Here is a quote from his "Universal Grammar", Theoria, 36: 373–398, 1970:

There is in my opinion no important theoretical difference between natural languages and the artificial languages of logicians; indeed I consider it possible to comprehend the syntax and semantics of both kinds of languages with a single natural and mathematically precise theory.

See, for instance, the entry on Montague Semantics in the Stanford Encyclopedia of Philosophy.

More broadly, the field of formal semantics (which Montague helped found) is dedicated to understanding and modelling natural language with the tools of logic and computer science.

A full mapping of natural language into formal logic is still a pipe dream, however. People are still debating how to understand quantifiers.

Title: Is there a homomorphic mapping between natural language and logic

Body: To me, this might be possible with a very complicated logic but then there should be no calculi that are useful. And is there an algorithm for it that could do that? This already looks very undecidable.</p>


问:
Prove a bounded subset of a compact set is closed
Consider R2 equipped with the Euclidean metric d2. Let A be a non-empty subset of R2, that is bounded and closed in (R2,d2). Let
S = {x^2 + y^2 : (x, y) ∈ A} ⊂ R

Question: Show that S is a closed subset of (R, | · |).

My attempt: I tried to prove by contradiction, that is, there exists an element x^2+y^2 in the closure of S but not in S. Then (x,y) is in the complement of A. Then there exists r>0 such that B((x,y),r) is in A complement. As x^2+y^2 is an accumulation point (limit point) of S, then for any ε>0, B(x^2+y^2)∩ S-{x^2+y^2} is not empty. I was stuck here and failed to define ε in terms of r

答:
Let $(r_n)$ be a sequence in $S$ which converges to some point $r$. We need to show that $r \in S$. (Let us denote $||(a,b)|| := \sqrt{a^2 + b^2}$, and let $K := ub(A)$)

As $r_n \in S \Rightarrow \exists (x_n,y_n) \in A$ s.t. $ ||(x_n,y_n)||^2 = r_n $.
As $A$ is closed, $\exists (x,y) \in A$ s.t. $(x_n,y_n) \rightarrow (x,y)$.

In order to prove the result, it suffices to show that $||(x_n,y_n)||^2 \rightarrow ||(x,y)||^2$ given $(x_n,y_n) \rightarrow (x,y)$. This follows directly from the triangle inequality and boundedness of $A$:
\begin{align*}
||(x_n,y_n)||^2 -  ||(x,y)||^2 &\leq (||(x_n,y_n)|| -  ||(x,y)||)(||(x_n,y_n)|| +  ||(x,y)||) \\
&\leq 2K(||(x_n,y_n)|| -  ||(x,y)||) \\
&\leq 2K||(x_n,y_n) - (x,y)||
\end{align*}
Qed.


问:
How to find all the prime and maximal ideals of $\mathbb R [x]$?
I know that $\mathbb{R}[x]$ is a PID. I have proven that all the ideals of the form $I=(x)$ are maximals ($\mathbb{R}[x]/(x)$ is isomorph to $\mathbb{R}$ which is a field. As a result I is maximal). I also proved that if $f(x)\in\mathbb{R}[x]$ is ireducable then $\mathbb{R}[x]/f(x)$ is isomorph to $\mathbb{C}$ which is a field, so $<f(x)>$ is maximal. Are those the only maximals ideals? How can I prove it? Are there any other prime ideals that are not maximals?

答:
The zero ideal is prime but not maximal. All the other prime ideals are maximal. Indeed, a nonzero prime ideal is generated by a prime element $f\in\mathbb{R}[x]$. But prime elements in an integral domain are irreducible, and so $f$ must be an irreducible polynomial. So now, if $(f)\subseteq (g)$ for some monic polynomial $g\in\mathbb{R}[x]$ then $g|f$, and since $f$ is irreducible it follows that $g=f$ or $g=1$. Thus $(g)=\mathbb{R}[x]$ or $(g)=(f)$. So indeed $(f)$ is a maximal ideal.

By the way, note that $f$ being irreducible doesn't imply that $\mathbb{R}[x]/(f)$ is isomorphic to $\mathbb{C}$. For example, if $f$ has degree $1$ then $\mathbb{R}[x]/(f)$ is isomorphic to $\mathbb{R}$. If $f$ is irreducible of degree $2$ then indeed $\mathbb{R}[x]/(f)\cong\mathbb{C}$.


问:
Can it be proved by contrapositive?: If $x | y^2$ then $x | y$
We want to show that if $x|y^2$ then $x|y$. Is this contrapositive proof correct?

Proof by contrapositive: Assume $x \not| \;y$ and $x,y \in Z$.  Thus, $y=xk+r$ for $k \in Z, r \in N, r < k$ by definition of not divides (there is a remainder). Hence, $y^2=(xk+r)^2=x^2k^2+2xk+r^2=x(xk^2+2k)+r^2$ for $r^2, xk^2+2k \in Z$. Therefore, $x \not| \;y^2 $ because there is a remainder $r^2$

答:
No, it is not correct. And it could not be, since the statement is false: $4\mid2^2$, but $4\nmid2$.

The error in your proof lies in assuming that $r^2$ is the remainder of the division of $y^2$ by $x$.


问:
Classification of prime ideals in $\mathbb{Z}[i]$
It's a couple of days that i'm struggling with this answer, which i'd like very much to understand.

I recall briefely what is the problem: I want to classify all prime ideals of $\mathbb{Z}[i]$.
The strategy is the following: for each (non-zero) prime ideal $(q)$ of $\mathbb{Z}$, we want to classify the prime ideals $Q$ of $\mathbb{Z}[i]$ that contain $(q)$. To do this, we first re-express the ring $Z[i]$ in a more convenient form, namely $\mathbb{Z}[i]\cong R:=\mathbb{Z}[x]/(x^2+1)$.  Now, the prime ideals of $R$ containing $qR$ are in order-preserving bijection with the prime ideals of $$R/qR=(\mathbb{Z}[x]/(x^2+1))/q(\mathbb{Z}[x]/(x^2+1))\cong \mathbb{Z}[x]/(q,x^2+1)\cong \mathbb{F}_q[x]/(x^2+1).$$

Then we consider various cases for the prime $q$ in $\mathbb{Z}$.

I've many question in all this, main questions are:

1) Consider for example, the case $q\equiv 1\bmod 4$. In this case $x^2+1$ is reducible , so that
$$R/qR\cong \mathbb{F}_q[x]/(x-a)(x+a)$$
which has two prime ideals, $(x-a)\mathbb{F}_q[x]/(x-a)(x+a)$ and $(x+a)\mathbb{F}_q[x]/(x-a)(x+a)$. Unwinding our various isomorphisms, this corresponds to a conjugate pair of Gaussian primes $\pi$, $\overline{\pi}$ with norm $q\equiv 1\bmod 4$ in $\mathbb{Z}[i]$. Well, i can't see how exactly to unwind our isomorphisms, who is $\pi$ concretely?

2)Secondly, admitting that $\pi$ and its conjugate $\overline{\pi}$ are the two Gaussian primes for this case, why do i multiply them? I mean, i have two primes in $R/qR$, but only one prime in $\mathbb{Z}[i]$, having norm $q$, why?

A different (last) question is: in this way i classify all ideals of $R$ such that their intersection with $\mathbb{Z}$ is a prime ideal in $\mathbb{Z}$. How can i know whether these ideals are all primes in $R$? I mean, if $P$ is a prime ideal of $R$, certainly $P\cap\mathbb{Z}$ is prime in $\mathbb{Z}$, but could an ideal $Q$ in $R$, not prime in $R$, have $Q\cap\mathbb{Z}$ prime in $\mathbb{Z}$?

Note: also a reference on this classification (not any classification of Gaussian prime ideals, but this particular proof) will be considered an answer.

答:
For your last question, every prime ideal of $\mathbb{Z}$ is contained in a maximal ideal of $R$, which is prime.

For the first question, notice that  $F_q [x]/(x^2+1)=(\mathbb{Z}[i])/(q)$. So since the first splits into a ring with two prime ideals, the second does as well, which shows that q is not prime, but in fact is composed of two primes. The ideal $(-x-a)=(x+a)$, so reversing the sign of $x$ switches between the ideals, which means the primes are complex conjugates. Finding the exact primes using this method does not work well; for instance, this method would suggest that the prime factors of 5 and 13 are $2+i,2-i$ and $5+i,5-i$ respectively (since 2 and 5 are the respective square roots of -1). The first case is true, and the second is not; the norm of $5-i$ is 26, not 13. The problem is that $5-i=(1-i)(2+3i)$, where $(2+3i)$ is the actual prime dividing 13.

So one method you could use to find the actual primes is to take $a+i$ (where $a$ is a root of -1 in your finite field), and find its norm; if its norm is not prime, that means you can factor $a+i$. Keep doing so till you get a piece with the right norm.


答: Let $I=(f)$ be an ideal of $\mathbb{R}[x]$; if $f$ is a constant non-zero (the ideal $(0)$ is not maximal) polynomial (i.e. it has degree $0$) then it is invertible; so $f$ must have degree $\geq1$. You have proved that, given that $f$ is irreducible, then $I$ is maximal (it is true indeed, but it is not true that the quotient is always isomorphic to $\mathbb{C}$). So, let us suppose this is not the case, and so $f(x)=p_1(x)\ldots p_n(x)$, with $p_i(x)$ non-constant and coprime with $p_j$, $j \neq i$; then the quotient $\mathbb{R}[x]/I \simeq \mathbb{R}[x]$ is not a field, since it can be splitted into a product of rings using the Chinese remainder Theorem for ideals.

Title: How to find all the prime and maximal ideals of $\mathbb R [x]$?

Body: I know that $\mathbb{R}[x]$ is a PID. I have proven that all the ideals of the form $I=(x)$ are maximals ($\mathbb{R}[x]/(x)$ is isomorph to $\mathbb{R}$ which is a field. As a result I is maximal). I also proved that if $f(x)\in\mathbb{R}[x]$ is ireducable then $\mathbb{R}[x]/f(x)$ is isomorph to $\mathbb{C}$ which is a field, so $<f(x)>$ is maximal. Are those the only maximals ideals? How can I prove it? Are there any other prime ideals that are not maximals?


问:
Prove that the normal curvature satisfies $\left|k_{n}\right| \geq 1 /|p|$
I'm taking a course in elementary differential geometry, and I was trying to understand a result.

Let $S$ be a nonempty, compact, oriented regular surface in $\mathbb{R}^{3}$ and let $p \in S$ be a point which maximises the function $f: S \rightarrow \mathbb{R}$ defined by $f(x)=|x|^{2}$. Prove that the normal curvature of any curve $C \subset S$ passing thought $p$ satisfies $\left|k_{n}\right| \geq 1 /|p|$.

Proof:
Let $\gamma:(-\epsilon, \epsilon) \rightarrow S$ be an arbitrary smooth curve parametrised by arc-length and $\gamma(0)=p .$ By the assumptions on $p,$ we must have $ \left.\frac{d}{d t}|\gamma(t)|^{2}\right|_{t=0}=0,\left.\quad \frac{d^{2}}{d t^{2}}|\gamma(t)|^{2}\right|_{t=0} \leq 0. $ Since above equations hold for all such $\gamma$, the first equation implies that the vector $p=\gamma(0)$ is proportional to the unit normal $N(p)=\pm \gamma(0) /|\gamma(0)|$ of $S$ at $\gamma(0)=0$. The second inequality implies that $ \left\langle\gamma^{\prime \prime}(0), \gamma(0)\right\rangle+\left\langle\gamma^{\prime}(0), \gamma^{\prime}(0)\right\rangle \leq 0 $, and...

Why is the normal given by $N(p)=\pm \gamma(0) /|\gamma(0)|$?

答: The first equation tells you that $p$ (as a vector) is always perpendicular to $\dot{\gamma}(0)$, whatever $\gamma$ you choose. Since all tangent vectors are of the form $\dot{\gamma}(0)$, then $p$ needs to be proportional to the normal vector.


问:
Chain rule notation when there is multivariables
Use the chain rule to determinate $u_s$ and $u_t$.

$u = x \csc(yz)$

$x = rs$

$y = s^2 t$

z =  $\frac{s}{t^2}$

Determinating $u_s$ means applying the chain rule using $t$ as a constant? How would be the proccess?

答:
$$\begin{align}
u_s&=x_s\csc(yz)+x\big(\csc(yz)\big)_s&\text{(product rule)}\\[1ex]
&=x_s\csc(yz)-x\csc(yz)\cot(yz)\big(yz\big)_s&\text{(chain rule)}\\[1ex]
&=x_s\csc(yz)-x\csc(yz)\cot(yz)(y_sz+yz_s)&\text{(product rule)}
\end{align}$$

The remaining derivatives are more immediate:

$$\begin{cases}x=rs\\y=s^2t\\z=\frac s{t^2}\end{cases}\implies\begin{cases}x_s=r\\y_s=2st\\z_s=\frac1{t^2}\end{cases}$$

Then the ultimate result follows by substituting everything in $u$:

$$\begin{align}
u_s&=r\csc\left(\frac{s^3}t\right)-rs\csc\left(\frac{s^3}t\right)\cot\left(\frac{s^3}t\right)\left(\frac{2s^2}t+\frac{s^2}t\right)\\[1ex]
&=r\csc\left(\frac{s^3}t\right)-\frac{3rs^3}t\csc\left(\frac{s^3}t\right)\cot\left(\frac{s^3}t\right)
\end{align}$$

which agrees with the derivative you added in the comments, but the final result should be free of the intermediate dependent variables $x,y,z$.


问:
Construction of a sequence of measurable function
Is there any sequence $f_n:\mathbb{R} \to \mathbb{R}$ of measurable functions such that for all $x \in \mathbb{R},\lim_nf_n(x)=x$ such that there exist a nonempty bounded interval $C_n=[\alpha_n,\beta_n]$ where $f_n$ is zero?

答:
For instance:$$f_n(x)=x\left(1-\mathsf1_{[0,\frac1{n}]}(x)\right)$$Addendum:

Another example is:$$f_n(x)=x\left(1-\mathsf1_{[n,e^n]}(x)\right)$$
with shifting interval increasing in length.


问:
$f'(x)=f(x)$ and $f(0)=0$ implies that $f(x)=0$ formal proof
How can I prove that if a function is such that $f'(x)=f(x)$ and also $f(0)=0$ then $f(x)=0$ for every $x$. I have an idea but it's too long, I want to know if there is a simple way to do it. Thanks!
Obviously in a formal way.</p>

答:
An implicit assumption is that the function is defined on some open interval containing $0$. Set

$$g(x)=e^{-x}f(x)$$

and compute the derivative:

$$g'(x)=-e^{-x}f(x)+e^{-x}f'(x)=-e^{-x}f(x)+e^{-x}f(x)=0$$

so the function $g$ is constant on the interval where it's defined. Since $g(0)=e^{-0}f(0)=0$ you can conclude that $g(x)=0$ for all $x$ and therefore also $f(x)=0$ for all $x$.

Without the initial assumption, you can get different functions with that property: define, for instance,
$$f(x)=\begin{cases}
0 & \text{if $-1<x<1$}\\
e^x & \text{if $2<x<3$}
\end{cases}$$
Then $f$ satisfies the requirements, but it's not constant.


问:
A Thousand Points
A 1000 points are graphed in the co-ordinate plane. Explain why it is possible to draw a straight line in the plane so that half the points on one side of the line and half are on the other.

This question was taken from Precalculus by Stewart,Watson.

答:
Consider the $xOy$ plane with $1000$ points randomly distributed. We want a line $y = mx+b$ such that above it we have $500$ points and under it we have the other $500$ points.

We start by choosing the slop of our line. We will choose the slop of our line in a such a way that this line is not parallel to any other line that passes through two of those $1000$ points.

Hence, this line will not intersect any of those $1000$ points or it will only intersect one of those points at a time.

Then, we choose a $b$ such that all the $1000$ points are under the line. By decreasing the value of $b$, the line will pass through the points one at time. In no moment, we will have two of those points belonging to this line at the same time.

So, by decreasing the line until it passes through $500$ points, we have that there are already $500$ points above it. Now, we have to make sure that the other $500$ are under it.

It may happen the case when we have $499$ points under it and the other one point belonging to the line. But we can surpass this by getting a difference $b$ such that this not happens.

This works because of the fact that between any two real numbers, there are a real number; something that I am not sure if you know, but by intuition you can see that is true.

Title: A Thousand Points


问:
A Function That Converge To The Min After Many Iterations Using Steepest Decent
I am searching for a function to show the "zig-zag" pattern of the steepest decent, I want to show that SGD converges after less iterations.

Which functions should look at?

答:
Steepest Descent (gradient descent with line search) exhibits zigzag behavior on the Rosenbrock Banana function.

$(x-1)^2 + 100(y-x^2)^2$ using the starting point (-1.2,1). Lots of zigzags to get to the optimum at (1,1). You can make it zigzag even more by increasing 100 to a larger number.